Local Pfaff Geometry

By openness choose $R>0$ with $\bar{B}(q_0,2R)\subset U$. Continuity of $X$ and $DX$ at $q_0$ gives finite $M,L\ge 0$ bounding $\mathrm{\parallel }X\mathrm{\parallel }$ on $\bar{B}(q_0,R)$ and $\mathrm{\parallel }DX\mathrm{\parallel }$ on $B(q_0,2R)$. By [F1] $X$ is totally differentiable there, and [F2] applied to the convex ball $B(q_0,2R)$ gives $\mathrm{\parallel }X(u)-X(v)\mathrm{\parallel }\le L\mathrm{\parallel }u-v\mathrm{\parallel }$ for $u,v\in \bar{B}(q_0,R)$.

Choose $\delta >0$ with $M\delta \le R\mathrm{/}2$ and $L\delta <1$, and let $\mathcal{C}:=\{\gamma \in C([-\delta ,\delta ],{\mathbb{R}}^m):\gamma (t)\in \bar{B}(q_0,R)\text{ for every }t\}$ with the supremum norm. It is complete: for a supremum-Cauchy sequence $({\gamma }_k)$, [F5] gives the pointwise limit $\gamma (t)\in {\mathbb{R}}^m$; the supremum-Cauchy estimate, passed to that pointwise limit, gives uniform convergence to $\gamma$. The limit is continuous: given $\epsilon >0$ and $t_0$, choose $k$ with $\mathrm{\parallel }{\gamma }_k-\gamma {\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon \mathrm{/}3$, then continuity gives $\rho >0$ such that $\mathrm{\parallel }{\gamma }_k(t)-{\gamma }_k(t_0)\mathrm{\parallel }<\epsilon \mathrm{/}3$ whenever $\mathrm{\mid }t-t_0\mathrm{\mid }<\rho$; the triangle inequality yields $\mathrm{\parallel }\gamma (t)-\gamma (t_0)\mathrm{\parallel }<\epsilon$. Also $\gamma (t)\in \bar{B}(q_0,R)$ for every $t$, since this Euclidean ball is closed and each ${\gamma }_k(t)$ lies in it. Thus $\gamma \in \mathcal{C}$. For $p\in \bar{B}(q_0,R\mathrm{/}2)$ define $T_p\gamma (t):=p+{\int }_0^{t}X(\gamma (s))ds$. The integrand is continuous and hence integrable by [F3]. Then $T_p$ maps $\mathcal{C}$ into itself, since $\mathrm{\parallel }{T}_p\gamma (t)-q_0\mathrm{\parallel }\le \mathrm{\parallel }p-q_0\mathrm{\parallel }+M\mathrm{\mid }t\mathrm{\mid }\le R$, and is a contraction with constant at most $L\delta$.

Starting at the constant curve ${\gamma }_0(t)=p$, iterate ${\gamma }_{k+1}=T_p{\gamma }_k$. If $\kappa =L\delta <1$, then $\mathrm{\parallel }{\gamma }_{k+1}-{\gamma }_k{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\kappa }^k\mathrm{\parallel }{\gamma }_1-{\gamma }_0{\mathrm{\parallel }}_{\mathrm{\infty }}$; the geometric tail bound makes $({\gamma }_k)$ uniformly Cauchy. Its limit $\varphi (\cdot ,p)\in \mathcal{C}$ is a fixed point by the contraction bound. Any two fixed points have distance at most $\kappa$ times that distance, so they coincide.

For $g(s)=X(\varphi (s,p))$, continuity and [F3] make every component Riemann integrable. At an interior $t$, the difference quotient of $\varphi (t,p)=p+{\int }_0^{t}g(s)ds$ differs from $g(t)$ by at most ${\sup }_{\mathrm{\mid }s-t\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }}\mathrm{\parallel }g(s)-g(t)\mathrm{\parallel }$, which tends to zero by continuity; thus ${\mathrm{\partial }}_t\varphi =X(\varphi )$. Conversely any $C^1$ solution staying in the ball satisfies the fixed-point equation by [F4], so uniqueness of the fixed point is uniqueness of the IVP.

If $p,p^{\mathrm{\prime }}$ lie in $\bar{B}(q_0,R\mathrm{/}2)$, subtract their fixed-point equations from step 3.1 and take suprema to obtain $\mathrm{\parallel }\varphi (\cdot ,p)-\varphi (\cdot ,p^{\mathrm{\prime }}){\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }p-p^{\mathrm{\prime }}\mathrm{\parallel }+L\delta \mathrm{\parallel }\varphi (\cdot ,p)-\varphi (\cdot ,p^{\mathrm{\prime }}){\mathrm{\parallel }}_{\mathrm{\infty }}$. Since $L\delta <1$, division gives the local Lipschitz estimate $(1-L\delta {)}^{-1}\mathrm{\parallel }p-p^{\mathrm{\prime }}\mathrm{\parallel }$.

Whenever the compositions remain in the chosen ball, both $t\mapsto \varphi (t+s,p)$ and $t\mapsto \varphi (t,\varphi (s,p))$ solve the same IVP at $t=0$, with initial value $\varphi (s,p)$. Uniqueness from step 4.1 gives the local flow law; setting $t=-s$ gives the local inverse relation.

For $m=0$, define ${\mathbb{R}}^0$ as the singleton consisting of the empty coordinate tuple, denoted $0$, with its unique vector-space structure and norm $\mathrm{\parallel }0\mathrm{\parallel }=0$. Its open subsets are empty or the singleton; the empty domain has no initial point and all assertions there are vacuous. On the singleton both fields necessarily equal zero. Define ${\varphi }_X^t(0)={\varphi }_Y^t(0)=0$ for every real $t$. These are the unique solution curves and satisfy the flow law. Every map into this zero vector space has the unique zero linear map as total derivative, since its differentiability remainder is identically zero; iterating gives continuous derivatives through order two. Thus the flows are jointly $C^2$, $J$ is the unique endomorphism of ${\mathbb{R}}^0$, which is also its identity $I_0$, and ${\mathrm{\partial }}_{t}J=DX({\varphi }_X)J=0$. The bracket and both transport expressions are zero, and both flow compositions are the identity. This proves every assertion for $m=0$ without invoking the positive-dimensional suppliers. In the remaining steps assume $m\ge 1$.

For $p$ near $q$ put ${\mathrm{\Delta }}_h(t):={\varphi }_X^t(p+h)-{\varphi }_X^t(p)$. The integral equations from [F1] and the one-variable fundamental theorem applied to $\theta \mapsto X({\varphi }_X^s(p)+\theta {\mathrm{\Delta }}_h(s))$ give ${\mathrm{\Delta }}_h(t)=h+{\int }_0^t{A}_h(s){\mathrm{\Delta }}_h(s)ds,A_h(s):={\int }_0^{1}DX({\varphi }_X^s(p)+\theta {\mathrm{\Delta }}_h(s))d\theta \mathrm{.}$ The segments lie in $K$ after shrinking the initial ball and time interval. The local Lipschitz estimate in [F1] gives ${\sup }_t\mathrm{\parallel }{\mathrm{\Delta }}_h(t)\mathrm{\parallel }=O(\mathrm{\parallel }h\mathrm{\parallel })$. Uniform continuity of $DX$ on $K$ from [F4] makes $A_h(s)$ converge uniformly to $DX({\varphi }_X^s(p))$ as $h\to 0$.

Let $\mathcal{I}=[-\delta ,\delta ]$ and let ${\mathcal{C}}_1=C(\mathcal{I},M_m(\mathbb{R}))$ with the supremum operator norm. This space is complete: for a supremum-Cauchy sequence $(J_k)$, each matrix entry at each $s$ is Cauchy because $\mathrm{\mid }{A}_{ij}\mathrm{\mid }\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$; [F5] applied to the $m^2$ entries gives a pointwise matrix limit $J(s)$, and the uniform Cauchy estimate passed to each entry gives uniform convergence of all entries. Each limit entry is continuous by the $\epsilon \mathrm{/}3$ argument for uniform limits of continuous functions. The finite-coordinate bound $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le {\sum }_{i,j}\mathrm{\mid }{A}_{ij}\mathrm{\mid }$ then gives uniform operator-norm convergence, so $J\in {\mathcal{C}}_1$. The space is nonempty because it contains the constant zero matrix curve. Put $A_0(s):=DX({\varphi }_X^s(p))$ and define $T(J)(t):=I_m+{\int }_0^t{A}_0(s)J(s)ds$. It maps ${\mathcal{C}}_1$ into itself and is a contraction with constant at most $L\delta <1$. By [F6], it has a unique fixed point $J$. Define $R_h(t):={\mathrm{\Delta }}_h(t)-J(t)h$. Subtracting the integral equations and writing the integrand difference as $(A_h-A_0){\mathrm{\Delta }}_h+A_0{R}_h$ gives $\mathrm{\parallel }{R}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le L\delta \mathrm{\parallel }{R}_h{\mathrm{\parallel }}_{\mathrm{\infty }}+\delta \mathrm{\parallel }{A}_h-A_0{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\mathrm{\Delta }}_h{\mathrm{\parallel }}_{\mathrm{\infty }}=o(\mathrm{\parallel }h\mathrm{\parallel })$, by step 1.2 and uniform continuity of $DX$ on $K$. Thus $D_q{\varphi }_X^t(p)=J(t,p)$. The continuous integrand in the equation for $J$ has average over $[t,t+h]$ tending to its value at $t$: the quotient error is at most ${\sup }_{\mathrm{\mid }s-t\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }}\mathrm{\parallel }DX({\varphi }_X^s(p))J(s,p)-DX({\varphi }_X^t(p))J(t,p)\mathrm{\parallel }$, which tends to zero. Thus ${\mathrm{\partial }}_{t}J=DX({\varphi }_X^t(p))J$ and $J(0,p)=I_m$.

Write $J_h(s)=D_q{\varphi }_X^s(p+h)$ and $J_0(s)=D_q{\varphi }_X^s(p)$. Subtracting their integral equations gives $J_h-J_0={\int }_0^t{A}_0(J_h-J_0)ds+{\int }_0^t(A_h-A_0)J_{h}ds$, where $A_h=DX({\varphi }_h)$, $A_0=DX({\varphi }_0)$, ${\varphi }_h={\varphi }_X^s(p+h)$, and ${\varphi }_0={\varphi }_X^s(p)$. The line-segment chain rule and Newton–Leibniz supplier claims in [F2]–[F3], applied entrywise to $DX$ along the segment from ${\varphi }_0$ to ${\varphi }_h$, give the exact formula $A_h-A_0={\int }_0^1{D}^{2}X({\varphi }_0+\theta ({\varphi }_h-{\varphi }_0))[{\varphi }_h-{\varphi }_0,\cdot ]d\theta$ and therefore $\mathrm{\parallel }{A}_h-A_0{\mathrm{\parallel }}_{\mathrm{\infty }}\le M_2\mathrm{\parallel }{\varphi }_h-{\varphi }_0{\mathrm{\parallel }}_{\mathrm{\infty }}$. By [F1], $\mathrm{\parallel }{\varphi }_h-{\varphi }_0{\mathrm{\parallel }}_{\mathrm{\infty }}=O(\mathrm{\parallel }h\mathrm{\parallel })$; the integral equation gives $\mathrm{\parallel }{J}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le (1-L\delta {)}^{-1}$. Absorbing the first integral using $L\delta <1$ yields $J_h-J_0=O(\mathrm{\parallel }h\mathrm{\parallel })$ uniformly in time. Uniform continuity of $D^{2}X$ on $K$, the same line-segment formula, and ${\varphi }_h-{\varphi }_0=J_{0}h+o(\mathrm{\parallel }h\mathrm{\parallel })$ from step 2.1 now give $A_h-A_0=D^{2}X({\varphi }_0)[J_{0}h,\cdot ]+o(\mathrm{\parallel }h\mathrm{\parallel })$ uniformly, and hence $(A_h-A_0)J_h=D^{2}X({\varphi }_0)[J_{0}h,J_0(\cdot )]+o(\mathrm{\parallel }h\mathrm{\parallel })$. For each base point $p$, define ${\mathcal{C}}_2=C(\mathcal{I},\mathrm{Bil}({\mathbb{R}}^m\times {\mathbb{R}}^m,{\mathbb{R}}^m))$ with the supremum bilinear-operator norm. The coordinatewise completeness proof of step 2.1 applies with $m^3$ coefficients, so ${\mathcal{C}}_2$ is complete; it is nonempty because it contains the zero bilinear-map curve. The map $(\mathcal{T}K)(t)[u,v]:={\int }_0^t(A_0(s)K(s)[u,v]+D^{2}X({\varphi }_0(s))[J_0(s)u,J_0(s)v])ds$ is a contraction with constant at most $L\delta$ on ${\mathcal{C}}_2$. By [F6] it has a unique fixed point $K_p$. Subtracting its equation evaluated at $(h,v)$ from the equation for $J_h-J_0$ gives $R_h:=J_h-J_0-K_p[h,\cdot ]$ with $\mathrm{\parallel }{R}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le L\delta \mathrm{\parallel }{R}_h{\mathrm{\parallel }}_{\mathrm{\infty }}+o(\mathrm{\parallel }h\mathrm{\parallel })$, so $R_h=o(\mathrm{\parallel }h\mathrm{\parallel })$. Therefore $D_{p}J[h](v)=K_p[h,v]$, and $K_p$ is the second derivative $D_q^2{\varphi }_X$. Its equation makes it bilinear. The coefficients, forcing, ${\varphi }_p$, and $J_p$ vary continuously in $p$ uniformly for $s\in \mathcal{I}$; for fixed-point maps $T_p$ with contraction constant $L\delta$, $\mathrm{\parallel }{K}_p-K_{p^{\mathrm{\prime }}}{\mathrm{\parallel }}_{\mathrm{\infty }}\le (1-L\delta {)}^{-1}\mathrm{\parallel }{T}_p(K_{p^{\mathrm{\prime }}})-T_{p^{\mathrm{\prime }}}(K_{p^{\mathrm{\prime }}}){\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. The integral equation gives continuity in $t$ as well. Thus the second derivatives in $q$ are continuous.

Let $V(t):=Y({\varphi }_X^t(q))$ and $W(t):=D_q{\varphi }_X^t(q)Y(q)$. By [F2] and the variational equation, $V^{\mathrm{\prime }}=DY({\varphi }_X^t)X({\varphi }_X^t)$ and $W^{\mathrm{\prime }}=DX({\varphi }_X^t)W$. The identity $[X,Y]=0$ makes $V$ solve this same linear equation with $V(0)=W(0)=Y(q)$; uniqueness from the contraction argument gives $D_q{\varphi }_X^t(q)Y(q)=Y({\varphi }_X^t(q))$.

On the open interior of the chosen time and initial-point neighborhood, step 2.1 gives $D_q{\varphi }_X=J$ and ${\mathrm{\partial }}_{t}J=DX({\varphi }_X)J$, while step 3.1 gives the continuous second derivative $D_{q}J=K$. The chain rule [F2] applied to ${\mathrm{\partial }}_t{\varphi }_X=X({\varphi }_X)$ gives ${\mathrm{\partial }}_{q_i}{\mathrm{\partial }}_t{\varphi }_X=DX({\varphi }_X)J{e}_i$, and the variational equation gives ${\mathrm{\partial }}_t{\mathrm{\partial }}_{q_i}{\varphi }_X=DX({\varphi }_X)J{e}_i$; also ${\mathrm{\partial }}_t^2{\varphi }_X=DX({\varphi }_X)X({\varphi }_X)$. Hence all first- and second-order coordinate partial derivatives of ${\varphi }_X$ exist and are continuous there. Apply [F7] pointwise first to ${\varphi }_X$ and then to its first-derivative map: its Jacobian is the matrix of first partials, and the Jacobian of that map is the continuous array of second partials. Thus ${\varphi }_X$ is jointly $C^2$.

For fixed $t$, the curve $s\mapsto {\varphi }_X^t({\varphi }_Y^s(q))$ has derivative $D{\varphi }_X^{t}Y({\varphi }_Y^s(q))=Y({\varphi }_X^t({\varphi }_Y^s(q)))$ by step 3.2, and starts at ${\varphi }_X^t(q)$. It is therefore the $Y$-flow from that initial point; IVP uniqueness from [F1] yields ${\varphi }_X^t{\varphi }_Y^s(q)={\varphi }_Y^s{\varphi }_X^t(q)$.

Fix $q\in U$ and distinct coordinates $i,j$. Choose $h,k>0$ small enough that the coordinate rectangle from $q$ to $q+h{e}_i+k{e}_j$ lies in $U$, and set $\mathrm{\Delta }=S(q+h{e}_i+k{e}_j)-S(q+h{e}_i)-S(q+k{e}_j)+S(q)$. By [F4] the second partials are uniformly continuous on the chosen closed rectangle. Applying [F2] to coordinate segments and [F3] to their continuous derivatives, Newton–Leibniz gives $\mathrm{\Delta }={\int }_0^h\text{\hspace{0.17em}⁣}{\int }_0^k{S}_{ij}(q+s{e}_i+t{e}_j)dtds={\int }_0^k\text{\hspace{0.17em}⁣}{\int }_0^h{S}_{ji}(q+s{e}_i+t{e}_j)dsdt\mathrm{.}$ Each equality follows by taking the two endpoint differences in the outer coordinate and then applying Newton–Leibniz to the inner difference; both integrands are continuous by $S\in C^2$.

Divide both expressions in step 1.1 by $hk$ and let $h,k\downarrow 0$. Continuity of $S_{ij}$ and $S_{ji}$ makes each average tend to its value at $q$, so $S_{ij}(q)=S_{ji}(q)$. Since $q,i,j$ were arbitrary, the coordinate formula in [F1] gives $d(dS)=0$.

Put $\lambda :=1\mathrm{/}\mu$. Total differentiability gives $\mu (q+h)=\mu (q)+D\mu (q)h+o(\mathrm{\parallel }h\mathrm{\parallel })$, and continuity with $\mu (q)\mathrm{\ne }0$ gives $\mu (q+h)\mu (q)=\mu (q{)}^2+o(1)$. Hence $\lambda (q+h)-\lambda (q)=-(\mu (q+h)-\mu (q))\mathrm{/}(\mu (q+h)\mu (q))=-\mu (q{)}^{-2}D\mu (q)h+o(\mathrm{\parallel }h\mathrm{\parallel })$, so $\lambda \in C^1(U)$ with continuous derivative $D\lambda =-{\mu }^{-2}D\mu$. For each $i$, $S_i={\mathrm{\partial }}_{i}S$ is $C^1$. The increment identity $\lambda (q+h)S_i(q+h)-\lambda (q)S_i(q)=\lambda (q+h)(S_i(q+h)-S_i(q))+S_i(q)(\lambda (q+h)-\lambda (q))$ and the total derivative expansions therefore give $D(\lambda S_i)=\lambda D{S}_i+S_{i}D\lambda$. Since $\omega =\lambda dS$, for coordinates $j<k$ the coefficient of $d{x}_j\wedge d{x}_k$ in $d\omega$ is ${\mathrm{\partial }}_j(\lambda S_k)-{\mathrm{\partial }}_k(\lambda S_j)=({\mathrm{\partial }}_j\lambda )S_k-({\mathrm{\partial }}_k\lambda )S_j$, as the mixed partials cancel by step 2.1. Thus $d\omega =d\lambda \wedge dS$, and $\omega \wedge d\omega =\lambda dS\wedge d\lambda \wedge dS=0$ because the alternating product vanishes when $dS$ occurs twice.

Since ${\omega }_{q_0}\mathrm{\ne }0$, one coordinate coefficient is nonzero at $q_0$; relabel that coordinate as $z$ and shrink the neighborhood so $b\mathrm{\ne }0$. Put $c_i:=-a_i\mathrm{/}b$ and $X_i:={\mathrm{\partial }}_{x_i}+c_i{\mathrm{\partial }}_z$. Then $X_i\in C^2$, $\omega (X_i)=a_i+b{c}_i=0$, and the $X_i$ are a basis of $\ker \omega$ because their first $n$ components form the standard basis.

Define $[X,Y]:=DYX-DXY$. Coordinate expansion of the displayed formula for $d\omega$ gives $d\omega (X,Y)=X[\omega (Y)]-Y[\omega (X)]-\omega ([X,Y])$. Thus for horizontal fields $d\omega (X_i,X_j)=-\omega ([X_i,X_j])$. Also $[X_i,X_j]$ has zero $x$-components, hence is vertical. Evaluating the three-form on $({\mathrm{\partial }}_z,X_i,X_j)$ gives $(\omega \wedge d\omega )({\mathrm{\partial }}_z,X_i,X_j)=bd\omega (X_i,X_j)$. Under (i), $d\omega (X_i,X_j)=0$; then $\omega ([X_i,X_j])=0$, and a vertical vector $v{\mathrm{\partial }}_z$ has $\omega (v{\mathrm{\partial }}_z)=vb$, so $[X_i,X_j]=0$.

By [F2]–[F3] and step 1.2, the fields $X_i$ have commuting local $C^2$ flows ${\varphi }_i^t$. Define $\mathrm{\Phi }(x,\eta ):={\varphi }_n^{x_n}\circ \cdots \circ {\varphi }_1^{x_1}(0,\eta )$ for $(x,\eta )$ near $(0,z_0)$. Each $X_i$ changes only $x_i$, at rate one, so the first $n$ coordinates of $\mathrm{\Phi }(x,\eta )$ are exactly $x$. Commutation and the transported-field identity in [F3] give ${\mathrm{\partial }}_{x_i}\mathrm{\Phi }=X_i(\mathrm{\Phi })$. At $(0,z_0)$ the columns of $D\mathrm{\Phi }$ are $X_1,\dots ,X_n,{\mathrm{\partial }}_z$; their first $n$ components form $I_n$ and the last column is ${\mathrm{\partial }}_z$, so $D\mathrm{\Phi }$ is invertible.

Let $N=n+1$. On an open product neighborhood of $(u_0,y_0)=((0,z_0),(0,z_0))$ where $\mathrm{\Phi }$ is defined, set $H(u,y):=\mathrm{\Phi }(u)-y$. Then $H(u_0,y_0)=0$ and its $u$-derivative $D_{u}H=D\mathrm{\Phi }(u_0)$ is invertible. Apply [L1] with ambient dimension $2N$ and target dimension $N$, permuting the input coordinates to $(y,u)$ so the $y$ block is free and the solved $u$ block has the invertible minor: conditionally, the zero set is the graph of a unique local $C^1$ map $G(y)$, so $\mathrm{\Phi }(G(y))=y$. Since $\mathrm{\Phi }$ is $C^2$, differentiating this identity by [F5] gives $D\mathrm{\Phi }(G(y))DG(y)=I$ and hence $DG(y)=(D\mathrm{\Phi }(G(y)){)}^{-1}$. The right side is $C^1$ (matrix inverse entries are cofactors divided by the nonzero determinant), so $G$ is $C^2$.

Let $S(x,z)$ be the last coordinate $\eta$ of $G(x,z)$. Then $S\in C^2$, $S(\mathrm{\Phi }(x,\eta ))=\eta$, and $S_z(0,z_0)=1$; after shrinking, $S_z\mathrm{\ne }0$. Differentiating along $X_i$ gives $S_{x_i}+c_i{S}_z=0$, so $dS=S_z(dz-{\sum }_i{c}_{i}d{x}_i)=(S_z\mathrm{/}b)\omega$. Thus $\mu :=S_z\mathrm{/}b$ is $C^1$ and nowhere zero. This proves (ii), conditionally on [L1].

Conversely, if (ii) holds, [F4] gives $\omega \wedge d\omega =0$ on that neighborhood, proving (i).