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The Lie algebra of a Lie group $G$ is the vector space $\mathfrak{g}=T_{e}G$ equipped with the bracket transferred from left-invariant vector fields. For $a\in G$, write $L_a(g)=ag$. A field is left-invariant when $(D_g{L}_a)X_g=X_{ag}$ for all $a,g$. Given $v\in T_{e}G$, defineThis is smooth and left-invariant, because $L_a{L}_g=L_{ag}$ and the chain rule applies. Conversely a left-invariant field is determined by its value at $e$, through this formula. Thusis a vector-space isomorphism, whose inverse is evaluation at the identity.

The Lie bracket of vector fields is preserved by diffeomorphisms: on functions this follows by transporting the commutator of derivations. Therefore the bracket of two left-invariant fields is again left-invariant. Define $[v,w]=[X_v,X_w{]}_e$. Bilinearity, antisymmetry and the Jacobi identity follow from the same properties of commutators of derivations. This supplies the asserted Lie algebra structure.

For the general linear group $G{L}_n(\mathbb{R})$, invertible matrices form an open subset of $M_n(\mathbb{R})$, so $T_{I}G{L}_n(\mathbb{R})=M_n(\mathbb{R})$. The left-invariant field determined by $A$ is $X_A(g)=gA$. Its ambient derivative is $D_g{X}_A(H)=HA$. Therefore differentiating left-invariant matrix fields givesHenceThis is the general linear Lie algebra, with the ordinary matrix commutator; right invariance with the same identification would give the opposite sign.

Now let $F:G\to H$ be a smooth Lie group homomorphism. It sends the identity to the identity and induces the linear mapSince $F\circ L_g=L_{F(g)}\circ F$, differentiation givesThese fields are $F$-related; no injectivity or surjectivity of $F$ is needed. For every smooth function $f$ on $H$,Apply this first with $w$ and then with $v$, and subtract the reversed order. The result isAt $e$, smooth functions detect tangent vectors, soThis proves that the differential of a Lie group homomorphism preserves Lie brackets, and hence that $F_{\ast }$ is a Lie algebra homomorphism.

For the exponential identity, let ${\eta }_v$ be the integral curve of $X_v$ through $e$. Left invariance and uniqueness show ${\eta }_v(t+s)={\eta }_v(t){\eta }_v(s)$ for small times: translating the curve through $e$ gives the curve through ${\eta }_v(s)$. Repeating this identity extends the curve to all real times. This proves completeness of left-invariant vector fields without assuming that an arbitrary smooth field is complete. Rescaling the parameter also gives ${\eta }_v(t)={\eta }_{tv}(1)$. By the definition of the Exponential map of a Lie group, ${\exp }_G(v)={\eta }_v(1)$.

The field-related identity shows that $F({\eta }_v(t))$ is an integral curve of $X_{F_{\ast }v}$ starting at $e_H$. By uniqueness it equals ${\eta }_{F_{\ast }v}(t)$ for all times. At time one,This is the naturality of the Lie group exponential.

Apply this to the determinant homomorphism $G{L}_n(\mathbb{R})\to {\mathbb{R}}^{\times }$. The derivative of the determinant at $I$ issince the permutation formula gives $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$. The left-invariant field on ${\mathbb{R}}^{\times }$ with initial tangent $c$ is $y\mapsto cy$; its curve through one solves $y^{\prime }=cy$, hence is $e^{ct}$. The matrix exponential determinant identity therefore follows from naturality:This argument uses only the flow definition of the exponential and the scalar exponential, not an unproved matrix formula.

To identify it explicitly with the usual matrix exponential, the series $E(t)={\sum }_{k\ge 0}{t}^k{A}^k/k!$ converges in operator norm, uniformly with its differentiated series on bounded intervals. Termwise differentiation gives $E^{\prime }=EA$ and $E(0)=I$. The series commutes with $A$, and differentiating $E(t)E(-t)$ gives zero, so $E(t)E(-t)=I$. Thus $E(t)$ stays in $G{L}_n(\mathbb{R})$ and, by uniqueness, is the integral curve of $X_A$ through $I$. The matrix series is exactly the flow-defined exponential, justifying either notation in the determinant identity.

Choose a generator $[S^n]$ of ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree of a continuous map is the integer determined byFor $n\ge 1$ this is ordinary top-dimensional homology; using reduced homology also gives a convention for $n=0$. Domain and target use the same orientation, so reversing the chosen generator in both leaves the integer unchanged. Homotopy invariance of homology makes the mapping degree invariant under homotopy.

We interpret the manifold criterion for a suspension in the category of nonempty topological manifolds without boundary, and “sphere” as homeomorphic to the standard sphere. Let $X$ have dimension $d$. If $X\cong S^d$, an explicit homeomorphism from its suspension to $S^{d+1}$ isIt is a continuous map that is a bijection: the two collapsed ends become the poles, and every other point determines $t$ and $x$ uniquely. The compact-to-Hausdorff continuous bijection theorem makes it a homeomorphism.

Conversely suppose $SX$ is a topological manifold. Its open part $X\times (0,1)$ shows its dimension is $d+1$. At a suspension vertex $v$ there is an open cone neighbourhood $U$ on $X$. It is contractible, while $U\setminus \{v\}$ admits a deformation retraction onto $X$. The long exact sequence in relative homology and excision give the local homology from a link calculationA point of a $(d+1)$-dimensional topological manifold without boundary has local homology $\mathbb{Z}$ in degree $d+1$ and zero in all other degrees. ConsequentlyThus $X$ is an integral homology sphere. This conclusion alone does not identify its homeomorphism type.

If $d\ge 2$, it is also simply connected. Choose a cone neighbourhood $U$ of the vertex and an open coordinate ball $B$ around the vertex inside $U$. By compactness of $X$, sufficiently short entire cone neighbourhoods fit inside any neighbourhood of $v$: the inverse image of that neighbourhood contains $X\times \{0\}$, and a finite subcover supplies one uniform collar length. Choose such a shorter cone $V\subset B$. The inclusions of punctured neighbourhoods factor asThe composite induces an isomorphism of fundamental groups, since both cone punctures are products of $X$ with intervals and their inclusion is a homotopy equivalence. The middle punctured ball has the homotopy type of $S^d$ and is simply connected for $d\ge 2$. Hence the composite is also zero on the fundamental group, forcing ${\pi }_1(X)=0$. This uses the nested neighbourhoods, not an unsupported claim that an arbitrary punctured neighbourhood is a sphere.

The Hurewicz theorem, applied successively to the vanishing lower homology groups, now gives ${\pi }_j(X)=0$ for $j<d$ and ${\pi }_d(X)\cong H_d(X)\cong \mathbb{Z}$. A representative of its generator is a continuous map $S^d\to X$ inducing an isomorphism in every integral homology degree. Topological manifolds have the homotopy type of CW complexes, so the homological Whitehead theorem makes it a homotopy equivalence. We have proved that $X$ is a homotopy sphere.

At this point the topological classification theorem is essential: the topological generalized Poincare theorem says that a closed homotopy sphere is homeomorphic to the standard sphere. Applying it completes the converse for $d\ge 2$. For $d=1$, the local calculation makes $X$ a connected compact one-dimensional manifold without boundary, hence a circle. For $d=0$, a compact zero-dimensional manifold is a finite set, and ${\tilde{H}}_0(X)\cong \mathbb{Z}$ forces exactly two points. Thus it is $S^0$. Altogether,The last classification step is not a consequence of homology alone, and we make no assertion of diffeomorphism with a prescribed smooth structure. If the word manifold allows boundary on both sides, the printed assertion needs this correction: the suspension of the interval is a closed disk. For instance $[x,t]\mapsto ((2x-1)(1-|2t-1|),2t-1)$ identifies it with the filled diamond $|u|+|v|\le 1$, although the interval is not a sphere.

To construct sphere maps of arbitrary integer degree, first use $h_m:S^1\to S^1$, $h_m(z)=z^m$ for any integer $m$, including the constant map when $m=0$. Lifting its argument to the real line gives $\theta \mapsto m\theta$, so the fundamental one-cycle is sent to $m$ times itself. Therefore $\mathrm{deg}{h}_m=m$.

Two cone neighbourhoods cover a suspension; their intersection admits a deformation retraction onto the original space. The reduced Mayer–Vietoris theorem gives a natural suspension isomorphismNaturality means that the suspension of a continuous map induces the same integer multiplier. This proves degree under suspension. Using the standard identification $S(S^k)\cong S^{k+1}$, setThus every integer occurs in every positive sphere dimension, with negative degrees handled just as well as positive ones.

If “every $n$” includes zero, that final printed assertion is false. The generator of ${\tilde{H}}_0(S^0)$ is the difference of its two points. A map of a two-point set is the identity, the transposition, or one of two constant maps, giving preciselyThese degrees of maps of the zero-sphere are the explicit exception to the arbitrary-degree construction.

The multiplicative norm first rules out zero divisors: if $xy=0$ and $x\ne 0$, then $0=|x||y|$ forces $y=0$. If the dimension is zero the claimed bound already holds, so suppose it is positive.

It is worth checking that the printed description of a ring with an underlying real vector space really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed $x$, the map $L_x(y)=xy$ is additive by distributivity, andIt is therefore continuous. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves $L_x(ty)=t{L}_x(y)$ for every real $t$. The same argument holds in the other variable. This is automatic real bilinearity from a multiplicative norm, so multiplication is a bilinear map between finite-dimensional vector spaces, and hence is smooth.

Let $S=\{x\in A:|x|=1\}$ be the unit sphere of the given positive-definite bilinear form. It is a copy of $S^{n-1}$. The squaring mapis well defined because $|x^2|=|x{|}^2=1$, and is smooth. Commutativity gives its derivativeFor $x\in S$ and $v\ne 0$, the norm identity gives $|D{Q}_x(v)|=2|v|$, so the derivative is injective, including on the tangent space $T_{x}S=x^{\perp }$. It takes tangent vectors into $T_{x^2}S$ since $Q$ maps the sphere to itself. Alternatively, polarization of $|xy{|}^2=|x{|}^2|y{|}^2$ gives $⟨xy,xz⟩=|x{|}^2⟨y,z⟩$, so $⟨x^2,2xv⟩=2⟨x,v⟩=0$ directly.

The only identifications under squaring are antipodal:The factorization uses commutativity, and the last implication uses the absence of zero divisors. Conversely $Q(-x)=Q(x)$. It follows that $Q$ induces an injective smooth mapThe antipodal quotient map is a local diffeomorphism, so the injectivity of $DQ$ implies injectivity of $Dq$. Since Real projective space is compact, the stated compact injective immersion is an embedding theorem makes $q$ a closed embedding.

For $n\ge 2$ its source and target have the same dimension $n-1$. The inverse function theorem makes its image open as well as closed in the connected sphere $S^{n-1}$. It is nonempty, so the image is the whole sphere; consequently $q$ is a diffeomorphism. But for $n\ge 3$ this is impossible:Indeed, the antipodal two-sheeted covering map $S^{n-1}\to {\mathbb{RP}}^{n-1}$ has simply connected total space in this range, and is its universal cover, with deck group $\mathbb{Z}/2$. A sphere of dimension at least two is simply connected. These fundamental groups cannot be those of homeomorphic spaces.

We have obtained the dimension bound for a commutative Euclidean normed algebra:The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map ${\mathbb{RP}}^1\to S^1$ is entirely compatible with the argument: the circle squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.

Under the given rank hypothesis, the necessary and sufficient local condition isVanishing only at the single point would not suffice: an extension preserves the coordinate Poisson brackets throughout its domain. Necessity follows from the coordinate criterion just proved.

For sufficiency, put $Q_i=a_i$ and $Y_i=X_{Q_i}$. Because the symplectic form is a nondegenerate bilinear form at every point, the independent differentials $d{Q}_i$ give independent fields $Y_1,\dots ,Y_n$. They commute, sinceMoreover $d{Q}_j(Y_i)=\{Q_j,Q_i\}=0$, so they span the tangent spaces of the $n$-dimensional fibres of the submersion $Q$. Choose a local section $s(Q)$ transverse to those fibres. Its joint local flow defines coordinates $(Q,t)$ byThe inverse function theorem makes $\Psi$ a local diffeomorphism, and commuting flows give ${\Psi }_{\ast }{\partial }_{t_i}=Y_i$.

In these coordinates ${\iota }_{{\partial }_{t_i}}{\Psi }^{\ast }{\omega }_0=d{Q}_i$. Each flow preserves the symplectic form, by Cartan's magic formula. Thus its coefficients are independent of $t$, andwhere $\beta =s^{\ast }{\omega }_0$ is a closed differential two-form on the base. On a small ball, the Poincaré lemma gives $\beta =d\alpha$, with $\alpha ={\sum }_i{\alpha }_i(Q)d{Q}_i$. DefineThenConsequently the original coordinates are extended to the required canonical transformation $(Q,P)$, with the prescribed first half $Q=a$. Its Jacobian matrix is invertible by construction. This is the canonical completion of independent commuting functions. For the printed $C^2$ data, the flows and resulting canonical transformation can be taken $C^1$, which is sufficient to preserve the symplectic form; smooth data give smooth coordinates.

Use the standard symplectic form ${\omega }_0={\sum }_{j=1}^{n}d{q}_j\wedge d{p}_j$. A canonical transformation is a diffeomorphism $f$ preserving this symplectic form:Thus, writing $f(q,p)=(Q,P)$, the defining identity is ${\sum }_{j}d{Q}_j\wedge d{P}_j={\sum }_{j}d{q}_j\wedge d{p}_j$. For a local canonical transformation, require this on its open domain. In particular preservation of the symplectic form, rather than just of phase-space volume, is the condition in dimensions greater than two.

Throughout these solutions choose the Hamiltonian vector field convention ${\iota }_{X_F}\omega =dF$. It gives $\dot{q}=F_p$, $\dot{p}=-F_q$ and agrees with the positive coordinate Poisson bracket used below. The opposite convention for Hamiltonian vector fields changes signs in the Lie bracket of vector fields identity but none of the commuting or integrability conclusions.

Fix a smaller angular strip and a momentum ball $B^{\prime }⋐B$; the canonical transformation will be defined there and map into the original domain for small $ϵ$. This domain restriction is natural for a near-identity change of coordinates with a nonzero momentum shift.

First solve the scalar cohomological equation on a Diophantine torusThe preceding result gives a real-analytic periodic $u$ on every smaller strip. We need a near-identity angular diffeomorphism $\chi$ and a constant vector $a$ such thatHere is a complete analytic construction of this translated conjugacy of a Diophantine vector field.

Write $w=ϵ\nabla u$ and, for a current approximation, putAssume $A$ is close to the identity. For a correction $\Delta \chi =Av$, the linearized defect isThe identity follows by differentiating the definition of $e$, so the term $(De)v$ already contains the current error. ChooseThe matrix average is invertible near the identity, and the right-hand side has zero mean. The preceding torus small-divisor estimate therefore solves this vector equation componentwise. Update ${\chi }_{+}=\chi +Av$, $a_{+}=a+\Delta a$. The exact new error isIt is quadratic in the current defect.

For convergence, reserve a fixed outer strip on which $w$ is analytic and work on shrinking inner strips whose total loss is less than half the starting width. Let $s=n+\tau$ and choose loss parameters ${\delta }_j=d{2}^{-j}$ with $0<d<1$ small enough that the fixed multiples of these losses used at each stage fit in the reserved total width. The torus small-divisor estimate and Cauchy estimates give, with constants uniform while $A$ stays close to the identity and $\chi$ remains in the reserved strip,andIndeed, the first remainder is bounded using $\Vert D{e}_j\Vert \le C{\delta }_j^{-1}\Vert e_j\Vert$, and the second using the uniformly bounded second derivatives of $w$ on the reserved strip. Taking a slightly larger exponent $\mu$ covers all these losses. Start with ${\chi }_0=\mathrm{id}$ and $a_0=0$, so $e_0=-w=O(ϵ)$. If a small constant $K$ satisfies $C{d}^{-\mu }K{2}^{2\mu }\le 1$ and $\Vert e_0\Vert \le K$, induction gives $\Vert e_j\Vert \le K{2}^{-2\mu j}$. The estimates on $\Delta {\chi }_j$ and $\Delta a_j$ are summable. Choosing $ϵ$ still smaller keeps the derivatives close to the identity and the images inside the reserved strip, closing the induction. The real periodic analytic limits $\chi ,a$ solve the conjugacy equation. Since $⟨w⟩=0$, the very first constant correction is zero; the first new defect is $O({ϵ}^2)$, and the remaining summed constant corrections are $O({ϵ}^2)$. Also the summed change of $\chi$ is $O(ϵ)$. This proves the claimed estimates as well as existence, without leaving an unsolved linear remainder.

Now set $\eta (x)=ϵ\nabla u(x)+a$ and define the symplectic cotangent lift with a closed momentum shiftThe cotangent lift of a diffeomorphism is symplectic. Translation by $\eta$ is also symplectic because the one-form $\eta \cdot dx=ϵdu+a\cdot dx$ is closed. The constant term need not be exact on the flat torus; closedness is sufficient. Thus their composition is the required near-identity canonical transformation.

Expanding the quadratic kinetic term gives a transformed linear coefficient $D{\chi }^{-1}(\omega +\eta \circ \chi )$, which is exactly $\omega$ by the conjugacy equation. The constant-in-$y^{\prime }$ part issince ${\mathcal{D}}_{\omega }u=⟨V⟩-V$. Consequently defineand, for $ϵ\ne 0$,In the exact identities, squared Euclidean norms mean the real polynomial ${\sum }_j{v}_j^2$, continued bilinearly on complex strips, without complex conjugation. The latter is uniformly bounded and analytic on the retained strip because $a=O({ϵ}^2)$. At $ϵ=0$ use the identity map, $E_0=0$, and $Q_0=|y^{\prime }{|}^2$. We have the exact requested normal formHere $E_{ϵ}=O(ϵ)$, $Q_{ϵ}$ is genuinely homogeneous quadratic in $y^{\prime }$, and its coefficient matrix differs from the identity by $O(ϵ)$. Thus $|Q_{ϵ}-|y^{\prime }{|}^2|\le C|ϵ||y^{\prime }{|}^2$. The remainder depends only on the angular coordinate. If the printed final $\tilde{V}(x)$ uses the old angle $x$, the same term is $\frac{1}{2}|\nabla u(x)+a/ϵ{|}^2$ evaluated at $x=\chi (x^{\prime })$; this is just the corresponding coordinate convention.

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.

The displayed estimate has two independent defects. Even for $a(v)=v$, its right side cannot use $f(t)$. At $t=1$, choose nonnegative smooth functions $\psi ,\chi$ with compact support and prescribe $f(1,x,v)=\psi (x)\chi (v)$. This is a legitimate transported solution with $f_{\mathrm{in}}(x,v)=\psi (x+v)\chi (v)$. The proposed inequality would requireVelocity dilation makes $\Vert \chi {\Vert }_1/\Vert \chi {\Vert }_{\infty }$ arbitrarily large, while the other factors stay fixed. Thus no universal $C_a$ works with the same-time mixed Lebesgue norm.

There is a second issue after replacing $f(t)$ by $f_{\mathrm{in}}$: the Jacobian determinant bounds give a local diffeomorphism, not necessarily a one-to-one map. Suppose additionally that $a$ is injective. With $y=x-ta(v)$, the change of variables formula and $|\det Da|\ge {\alpha }_1$ giveSurjectivity is not needed because the domain of integration can be enlarged. With injectivity, the corrected estimate has $C_a=1/{\alpha }_1$. The upper bound ${\alpha }_2$ is unnecessary.

More generally, dispersion with a nonlinear velocity map uses the area formula to sum over inverse branches. Define the weighted inverse multiplicityIf ${\mathrm{esssup}}_w{m}_a(w)<\infty$, the same argument gives the corrected estimate with $C_a=\Vert m_a{\Vert }_{\infty }$. At most $N$ inverse branches give $C_a\le N/{\alpha }_1$. Thus the missing global assumption concerns multiplicity, not merely local volume distortion.

For a noninjective map with constant Jacobian determinant giving a counterexample in two dimensions, write $v=(s,r)$ and setThe polar coordinates calculation gives $\det Da=e^{2s}{e}^{-2s}=1$, yet $a(0,2\pi k)=(1,0)$ for every integer $k$. To turn this into a failure of the estimate, take a small open disk $W$ about $(1,0)$ that avoids the origin. For each of $N$ inverse branches over $W$, choose a smooth cutoff ${\chi }_k(v)=\chi (a(v))$ supported on that branch, where $\chi$ has support strictly inside $W$ and is extended by zero. These velocity supports are disjoint. SetThen $\Vert f_{\mathrm{in}}{\Vert }_{L_x^1{L}_v^{\infty }}=\Vert \psi {\Vert }_1\Vert \chi {\Vert }_{\infty }$, independent of $N$, whereas at a fixed $t>0$ and suitable $x$,The last integral is positive and independent of $N$. No finite $C_a$ exists for this fixed $a$, although ${\alpha }_1={\alpha }_2=1$. In one dimension, by contrast, a nonvanishing derivative has a constant sign, so the lower derivative bound makes $a$ a global diffeomorphism.

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.

An $n$-dimensional smooth manifold is a Hausdorff space with a countable topological base, equipped with a smooth atlas of homeomorphisms from open subsets onto open subsets of ${\mathbb{R}}^n$, whose overlap maps are smooth diffeomorphisms. The smooth structure is the maximal smooth atlas compatible with these manifold charts. Here manifolds have no boundary unless specified otherwise.

For a space with all the requested topological properties but no such atlas, take the topological tripod: three closed intervals joined at one endpoint. It is a compact connected subspace of the plane with its induced metric. A countable base of planar rational balls restricts to a countable base, the metric makes it Hausdorff, and compactness gives a finite subcover of every open cover, hence a locally finite refinement and paracompactness.

At an interior point of any arm, arbitrarily small neighborhoods are intervals. A coordinate ball in dimension at least two would remain connected after deleting its center; an interval does not. Dimension zero would make the space discrete. Thus any possible connected manifold structure would have dimension one. But a sufficiently small neighborhood of the junction, minus the junction, has three components, whereas an interval chart has two. This contradiction rules out even a topological manifold structure, and hence any smooth one.

The product smooth structure uses manifold charts $(\varphi ,\psi ):U\times V\to \varphi (U)\times \psi (V)\subset {\mathbb{R}}^{m+n}$. Transition maps act separately in the two coordinate blocks and are smooth with smooth inverses. Products of countable bases give a countable base, and the product remains Hausdorff. Therefore $\boxed{\dim (M\times N)=\dim M+\dim N}$ with this natural smooth structure.

Define the tangent space by point derivations: a tangent vector at $x$ is an $\mathbb{R}$-linear map $D$ on germs of smooth functions at $x$, satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. Addition and scalar multiplication preserve this rule, so these derivations form a vector space. In coordinates $u^1,\dots ,u^n$, the local identityfollows by integrating the derivative of $h$ along the coordinate line segment. Derivations annihilate constants, so it gives $Dh={\sum }_{i}D(u^i){\partial }_{i}h(u(x))$. The coordinate derivations ${\partial }_i{|}_x$ are independent since they evaluate the coordinate functions as ${\delta }_i^j$. ThusFor a smooth map, define the differential of a smooth map intrinsically by $(f_{\ast }D)(h)=D(h\circ f)$. It is again linear and satisfies the derivation rule at $f(x)$, so it maps $T_{x}M$ into $T_{f(x)}N$. Its coordinate matrix is the Jacobian of the coordinate expression of $f$, independently of the charts by the intrinsic definition and chain rule.

Finally, a zero differential forces each target coordinate function to have all derivatives zero on a small connected source coordinate ball mapping into one target chart. Integration on straight segments makes those functions constant there. Hence $f$ is locally constant. Each nonempty fiber is both open and closed, so connectedness gives the zero-differential constancy theorem, $\boxed{f\text{ is constant}}$.

For a diffeomorphism, the differential and its inverse preserve the natural pairing of a vector with a covector. Consequently the mixed tensor pullback commutes with every tensor contraction $C$:Differentiating at zero gives ${\mathcal{L}}_X(CT)=C({\mathcal{L}}_{X}T)$. The mixed tensor pullback also preserves tensor products, soThe ordinary product rule for differentiation therefore yieldsThis proves the contraction and Leibniz rule properties for all smooth generators, without needing straightening coordinates.

The suggested flow-box theorem applies locally only where $X\ne 0$. For example $X=x{\partial }_x$ vanishes at $x=0$, so it cannot equal a coordinate basis vector there. Its local flow is nevertheless ${\varphi }_s(x)=e^{s}x$ and ${\varphi }_s^{\ast }dx=e^{s}dx$, giving ${\mathcal{L}}_{X}dx=dx$ even at that zero. This Lie derivative at a zero of its generator illustrates why the general flow proof is needed to cover every point.

The pure gravitational action is invariant under diffeomorphisms. An infinitesimal diffeomorphism generated by a compactly supported vector field $\xi$ changes the metric by its tensor Lie derivative,Using symmetry of $E^{ab}$ and the variational expression already derived,As ${\xi }_b$ is arbitrary, the off-shell metric divergence identity isThis Noether identity follows for every metric tensor without imposing either the gravitational field equation or the matter equations. It is a consequence of the metric action's diffeomorphism invariance, not a conclusion requiring a long component calculation.

The shear $(t,x,v)\mapsto (t,x+tv,v)$ is a smooth diffeomorphism with determinant one and sends compact sets to compact sets. Thus $F,G$ are locally integrable functions. On a compact time interval $I$ and compact label set $K\subset {\mathbb{R}}^6$, the integral ${\int }_K{\int }_I|G|dtdxdv$ is finite. The Fubini theorem, followed by a countable exhaustion of time intervals and label sets, shows that $G(\cdot ,x,v)\in L_{\mathrm{loc}}^1(\mathbb{R})$ for almost every $(x,v)$.

To transform the weak equation, choose $\varphi (t,x,v)=\rho (x-tv,v)\psi (t)$, where $\rho \in C_c^{\infty }({\mathbb{R}}^6)$ and $\psi \in C_c^{\infty }(\mathbb{R})$. Including a velocity cutoff in $\rho$ ensures the required compact support. The transport derivative of this test function is $\rho (x-tv,v){\psi }^{\prime }(t)$. Change variables to the characteristic labels to obtainThe fundamental test-function identity makes the bracket zero almost everywhere. Choose a countable dense family of time test functions on each bounded interval and extend by continuity; there is then a single negligible label set outside which ${\partial }_{t}F=G$ as one-dimensional distributions.

For each remaining label, subtract the primitive $H(t)={\int }_0^{t}G(s)ds$. The distributional derivative of $F-H$ is zero, so it is a constant almost everywhere. Choose the resulting absolutely continuous representative along free characteristics, $\tilde{F}(t)=C+H(t)$. The fundamental theorem of calculus givesThe representative qualification is essential: an arbitrary locally integrable representative can be modified on one time slice without changing the distributional equation, and would not satisfy the identity for every pair of times.

Moser's trick turns variation of symplectic forms into an equation for a time-dependent vector field. Let ${\omega }_t$, $0\le t\le 1$, be a smooth path of symplectic forms on a compact manifold without boundary, with a constant de Rham cohomology class. Choose a smooth family of one-forms ${\alpha }_t$ such that ${\dot{\omega }}_t=d{\alpha }_t$. Such a smooth choice can be made using a fixed auxiliary metric; the essential requirement is this exactness throughout the path.

Nondegeneracy uniquely determines $X_t$ byLet $f_t$ be its flow, with $f_0=\mathrm{id}$. Compactness ensures existence over the whole parameter interval. By Cartan's magic formula,ThusThe path is made constant by a diffeomorphism moving with $X_t$. Every interpolating form must be nondegenerate; equal endpoint cohomology alone does not ensure that every linearly interpolated form is a symplectic form. On a noncompact manifold one instead needs completeness of this flow, or restricts to a sufficiently small neighborhood, as in the local argument below.