Darboux's Theorem: Local Canonical Coordinates

Derives Corollary A12.3. Let \(u\) and \(u'\) be canonical charts around \(P\) and \(P'\) furnished by Theorem A12.2, both with image containing a ball \(B\subset\R^{2m}\) about the origin. Put \(f = (u')^{-1}\circ u\) on \(U = u^{-1}(B)\). Both \(\omega\) and \(\omega'\) read as the same constant form Equation (A12.1) in their own charts, and the pullback of a form is computed chartwise (Definition A12.4 below), so \(f^{\ast}\omega' = \omega\). Any local invariant would have to take the same value at \(P\) and \(P'\), and \(\omega\), \(\omega'\) were arbitrary.

Derives Lemma A12.5. Composition. Equation (A12.6) is the chain rule Equation (11.91) inserted \(k\) times into Equation (A12.3): \(\pp_{i}\left(g\circ f\right)^{j} = \pp_{l}g^{j}(f(x))\,\pp_{i}f^{l}(x)\).

Scalars and their differentials. For \(k=0\), \(f^{\ast}g=g\circ f\) and \(\pp_{i}(g\circ f) = \pp_{j}g(f(x))\,\pp_{i}f^{j}\), which is Equation (A12.3) for the one-form \(\dd g\); hence

Wedge. It suffices to check Equation (A12.4) on the components. Both sides are multilinear in \(\alpha\) and \(\beta\), so it is enough to take \(\alpha = \dd y^{a_{1}}\wedge\cdots\wedge\dd y^{a_{k}}\) and \(\beta = \dd y^{b_{1}}\wedge\cdots\wedge\dd y^{b_{l}}\) with the \(y\) the coordinates on \(V\). By Definition 17.102 and Equation (A12.3), the pullback of a wedge of coordinate differentials is the wedge of their pullbacks: for two of them,

and the same cancellation of index pairs, term by term over the permutations, gives \(f^{\ast}\left(\dd y^{a_{1}}\wedge\cdots\wedge\dd y^{a_{k}}\right) = \dd f^{a_{1}}\wedge\cdots\wedge\dd f^{a_{k}}\), where \(\dd f^{a} = f^{\ast}\dd y^{a}\) by Equation (A12.7). Equation (A12.4) follows because both sides then read \(\dd f^{a_{1}}\wedge\cdots\wedge\dd f^{a_{k}}\wedge \dd f^{b_{1}}\wedge\cdots\wedge\dd f^{b_{l}}\).

Exterior derivative. Write \(\alpha = \frac{1}{k!}\alpha_{a_{1}\ldots a_{k}}\, \dd y^{a_{1}}\wedge\cdots\wedge\dd y^{a_{k}}\). By the two paragraphs just proved,

Applying \(\dd\) and using the Leibniz rule Equation (17.246) repeatedly, every term in which \(\dd\) falls on some \(\dd f^{a}\) vanishes by nilpotency Equation (17.245), so

the last equality by Equation (A12.7) applied to the component functions and one more use of the two paragraphs above.

Derives Lemma A12.6. Let \(S=\max_{[0,T]}u\), finite by Theorem 11.24. We show by induction that for every \(N\ge0\)

For \(N=0\), Equation (A12.8) and \(u\le S\) give \(u(t)\le c + LSt\). Assuming Equation (A12.9) at \(N\) and substituting it into Equation (A12.8), term-by-term integration gives

which is the statement at \(N+1\). The last term of Equation (A12.9) tends to \(0\) as \(N\to\infty\), because the exponential series converges (Definition 11.53), and the sum tends to \(\ee^{Lt}\).

Derives Lemma A12.7. Fix \(x\in K\) and \(r>0\) with \(\overline{B}_{r}(x)\subset K\). For \(0<\abs{s}\le r\) the mean value theorem Theorem 11.35 applied to \(\lambda\mapsto F(t,x+\lambda\vect{e}_{i})\) gives \(\theta = \theta(t,s)\in(0,1)\) with

The set \([0,1]\times\overline{B}_{r}(x)\) is compact, so \(\pp_{i}F\) is uniformly continuous there (Theorem 11.25): given \(\varepsilon>0\) there is \(\delta>0\) with \(\abs{\pp_{i}F(t,y)-\pp_{i}F(t,x)}<\varepsilon\) whenever \(\abs{y-x}<\delta\), uniformly in \(t\). Hence for \(\abs{s}<\delta\) the difference quotient of \(G\) differs from \(\int_{0}^{1}\pp_{i}F(t,x)\dd t\) by at most \(\varepsilon\), which is Equation (A12.10). Continuity of the right-hand side in \(x\) follows from the same uniform continuity.

Derives Theorem A12.8. A Lipschitz box. Choose \(r>0\) with \(\overline{B}_{2r}(p)\subset U\) and put \(Q = [0,1]\times\overline{B}_{2r}(p)\), a compact set. On \(Q\) the continuous functions \(X\) and \(\pp_{j}X^{i}\) are bounded (Theorem 11.24); set \(M=\max_{Q}\abs{X}\) and

with \(D_{x}X\) the matrix \(\pp_{j}X^{i}\) and \(\norm{\cdot}\) as in Equation (A12.2). Since \(\overline{B}_{2r}(p)\) is convex, Theorem 11.35 applied componentwise along the segment from \(z\) to \(y\) bounds each component of the difference by \(\abs{\nabla_{x}X^{i}}\,\abs{y-z}\le\norm{D_{x}X}\abs{y-z}\), and summing the \(n\) squares gives

Existence on all of \([0,1]\). The constant curve \(y\equiv p\) solves Equation (A12.11) with \(y(0)=p\), because \(X_{t}(p)=0\); by uniqueness in Theorem 13.8 it is the solution through \(p\), which is the third assertion of Equation (A12.11). Let now \(x\in B := B_{\delta}(p)\) with \(\delta = r\,\ee^{-L}\), and let \(\psi_{t}(x)\) be the maximal solution. So long as it remains in \(\overline{B}_{2r}(p)\), the integral form of the equation and Equation (A12.13) give

so Lemma A12.6 yields

The solution therefore never approaches the boundary of \(\overline{B}_{2r}(p)\). Since the Picard step \(h=\min\set{1,\;r/M}\) of Theorem 13.8 is the same at every starting time in \([0,1]\) and every starting point of \(\overline{B}_{r}(p)\), the solution can be continued in steps of that fixed length and reaches \(t=1\) after finitely many of them.

Lipschitz dependence. For \(x,y\in B\) the same computation with \(p\) replaced by \(y\), using Equation (A12.13) (both curves stay in \(\overline{B}_{r}(p)\) by Equation (A12.14)) and Lemma A12.6, gives

The variational equation. Put \(A(t,x) = D_{x}X\bigl(t,\psi_{t}(x)\bigr)\), continuous on \([0,1]\times B\) with \(\norm{A}\le L\), and let \(\Phi(\cdot,x)\) be the unique solution of the linear system

which exists on the whole of \([0,1]\) by Corollary 13.9. Its integral form and Lemma A12.6 give \(\norm{\Phi(t,x)}\le\sqrt{n}\,\ee^{L}\).

\(\Phi\) is continuous in \(x\). Subtracting the integral forms of Equation (A12.16) at \(x\) and at \(y\),

The second integral is bounded by \(\sqrt{n}\,\ee^{L}\,\eta(x,y)\) with \(\eta(x,y)=\max_{s}\norm{A(s,x)-A(s,y)}\), and \(\eta(x,y)\to0\) as \(y\to x\): \(D_{x}X\) is uniformly continuous on the compact \(Q\) (Theorem 11.25) and \(\abs{\psi_{s}(x)-\psi_{s}(y)}\le\ee^{L}\abs{x-y}\) by Equation (A12.15). Lemma A12.6 then gives \(\norm{\Phi(t,x)-\Phi(t,y)}\le\sqrt{n}\,\ee^{2L}\eta(x,y)\).

\(\Phi\) is the derivative. Fix \(x\in B\) and let \(h\) be small enough that \(x+h\in B\). Put \(\Delta(t) = \psi_{t}(x+h)-\psi_{t}(x)\) and \(R(t) = \Delta(t)-\Phi(t,x)h\), so \(R(0)=0\) and

where \(\rho(t) = X_{t}(z+\Delta)-X_{t}(z)-D_{x}X(t,z)\Delta\) with \(z=\psi_{t}(x)\). Applying Theorem 11.35 to \(\lambda\mapsto X^{i}_{t}(z+\lambda\Delta)\) gives \(\theta_{i}\in(0,1)\) with \(\rho^{i}(t) = \left[\pp_{j}X^{i}(t,z+\theta_{i}\Delta) -\pp_{j}X^{i}(t,z)\right]\Delta^{j}\), whence \(\abs{\rho(t)}\le w\bigl(\abs{\Delta}\bigr)\abs{\Delta}\), where \(w\) is \(\sqrt{n}\) times a modulus of uniform continuity of \(D_{x}X\) on the compact set \(Q\) (Theorem 11.25), so that \(w(\lambda)\to0\) as \(\lambda\to0\). With \(\abs{\Delta}\le\ee^{L}\abs{h}\) from Equation (A12.15), the integral form of the equation for \(R\) and Lemma A12.6 give

so \(\abs{R(t)}/\abs{h}\to0\) as \(h\to0\). By Definition 11.99, \(\psi_{t}\) is differentiable at \(x\) with differential \(\Phi(t,x)\); the previous paragraph makes the partial derivatives continuous, so \(\psi_{t}\) is \(C^{1}\) (Definition 11.98).

Derivatives of every order. We show by induction on \(k\ge1\): if \(X\) is \(C^{k}\) then \(\psi\) is \(C^{k}\) in \(x\). The case \(k=1\) is what was just proved. Let \(X\) be \(C^{k+1}\) and consider the enlarged system on \(U\times\R^{n\times n}\),

whose right-hand side is \(C^{k}\) in \((y,\Phi)\) because \(X\) is \(C^{k+1}\). Its solution is exactly \(\bigl(\psi(t,x),\Phi(t,x)\bigr)\), and the induction hypothesis applied to Equation (A12.17) makes it \(C^{k}\) in the initial data, hence in \(x\). So \(\pp\psi/\pp x = \Phi\) is \(C^{k}\) in \(x\), i.e.\ \(\psi\) is \(C^{k+1}\) in \(x\). Since moreover \(\pp_{t}\psi = X_{t}(\psi)\), every derivative in \(t\) is a polynomial expression in derivatives of \(X\) and of \(\psi\) in \(x\); hence all mixed partial derivatives of \(\psi\) in \((t,x)\) of every order exist and are continuous, and \(\psi\) is smooth.

Diffeomorphism. Fix \(t\in[0,1]\) and let \(\eta_{s}\) be the flow built by the same construction for the field \(Y_{s}(y) = -X_{t-s}(y)\), which also vanishes at \(p\); let \(B_{2}\) be its ball. That ball is the same for every \(t\): the construction used only \(r\), \(M\) and \(L\), and those depend on \(X\) and the box \(Q\) alone, not on \(t\), while \(\abs{Y}\) and \(\norm{D_{x}Y}\) have the same maxima over \(Q\) as \(\abs{X}\) and \(\norm{D_{x}X}\). Shrinking \(B\) once and for all so that \(\ee^{L}\delta\) is smaller than the radius of \(B_{2}\), Equation (A12.14) puts \(\psi_{t}(B)\) inside \(B_{2}\) for every \(t\). For \(x\in B\) the curve \(s\mapsto\psi_{t-s}(x)\) satisfies \(\dd/\dd s = -X_{t-s}(\psi_{t-s}(x)) = Y_{s}(\cdot)\) and starts at \(\psi_{t}(x)\), so uniqueness (Theorem 13.8) gives \(\eta_{t}(\psi_{t}(x)) = \psi_{0}(x) = x\). Put \(W = \set{y\in B_{2}\mid \eta_{t}(y)\in B}\), open because \(\eta_{t}\) is continuous. Then \(\psi_{t}(B)\subseteq W\); and for \(y\in W\) the same uniqueness argument run backwards gives \(\psi_{t}(\eta_{t}(y)) = y\), so \(W\subseteq\psi_{t}(B)\). Hence \(\psi_{t}:B\longrightarrow W\) is a bijection onto an open set, with inverse \(\eta_{t}\), and both are smooth.

Derives Lemma A12.9. By Equation (A12.3),

Differentiate in \(t\) by the product rule. The first factor contributes, by the chain rule Equation (11.91) and Equation (A12.18),

For the remaining \(k\) factors, the mixed partial derivatives of \(\varphi\) may be exchanged (Proposition 11.105), so

again by Equation (11.91). The \(a\)-th such term is therefore

which, after renaming the summation indices \(j_{a}\leftrightarrow l\), is the pullback of the \(a\)-th drag term of Equation (A12.20). Collecting the \(k+1\) contributions gives Equation (A12.19). Finally, taking \(\alpha_{t}=\alpha\) independent of \(t\), \(\varphi_{0}=\id\) and \(X_{t}=X\), the identity at \(t=0\) reads \(\dd\left(\varphi_{t}^{\ast}\alpha\right)/\dd t\big|_{0} =\mathcal{L}_{X}\alpha\) with the right-hand side Equation (A12.20): this is the definition Equation (17.267) of the Lie derivative, so Equation (A12.20) is indeed the \(k\)-form case of Proposition 17.127 — one transport term and one drag term per index — and no separate verification is needed.

Derives Lemma A12.11. Translating, take \(x_{0}=0\). Put \(\varphi_{t}(x)=tx\) for \(t\in(0,1]\), which maps \(B\) into \(B\) by Definition A12.10, and

which is Equation (A12.18). Both \(\varphi\) and \(X\) are smooth on \((0,1]\times B\). By Lemma A12.9 with \(\alpha_{t}=\alpha\) fixed, and by Cartan's magic formula Equation (17.276) together with \(\dd\alpha=0\),

the last step by Equation (A12.5). The components of the form being differentiated are

since \(\left(i_{X_{t}}\alpha\right)_{i_{2}\ldots i_{k}}(y) = t^{-1}y^{j}\alpha_{j\,i_{2}\ldots i_{k}}(y)\) by the interior-product convention of Proposition 17.128, each of the \(k-1\) pullback factors \(\pp_{i}\varphi^{j}_{t}=t\,\delta^{j}_{\ i}\) contributes a \(t\), and \(y=tx\) supplies \(y^{j}/t = x^{j}\). The right-hand side of Equation (A12.23) extends continuously to \(t=0\) because \(k\ge1\), and so do its partial derivatives in \(x\), so \(\sigma\) of Equation (A12.21) is a well-defined smooth \((k-1)\)-form and, by Lemma A12.7, \(\dd\) may be taken under the integral sign. Integrating Equation (A12.22) from \(\varepsilon\) to \(1\),

Now \(\varphi_{1}=\id\), so the first term is \(\alpha\); and \(\left(\varphi_{\varepsilon}^{\ast}\alpha\right)_{i_{1}\ldots i_{k}}(x) = \varepsilon^{k}\alpha_{i_{1}\ldots i_{k}}(\varepsilon x)\) tends to \(0\) as \(\varepsilon\to0\), again because \(k\ge1\) and \(\alpha\) is continuous at the origin. Letting \(\varepsilon\to0\) in Equation (A12.24), with the interchange of \(\dd\) and the integral justified by Lemma A12.7, gives \(\alpha=\dd\sigma\) with \(\sigma\) as in Equation (A12.21). Setting \(x=x_{0}\) there makes the factor \((x-x_{0})^{j}\) vanish, so \(\sigma(x_{0})=0\).

Proof of Theorem A12.2. Derives Theorem A12.2. Step 1: the pointwise normal form. Let \(n=\dim M\). The bilinear form \(\omega_{P}\) on the \(n\)-dimensional real vector space \(T_{M}(P)\) is antisymmetric and non-degenerate (Definition A12.1), so Proposition 9.119 applies: \(n=2m\) is even and there is a basis \(\set{e_{1},\ldots,e_{m},f_{1},\ldots,f_{m}}\) of \(T_{M}(P)\) in which Equation (9.143) holds, that is, in which the matrix of \(\omega_{P}\) is Equation (9.144).

Step 2: a chart adapted at the point. Choose any chart \(x=(x^{1},\ldots,x^{2m})\) around \(P\) with \(x(P)=0\) and with the coordinate vectors \(\pp_{i}\) at \(P\) equal to that basis — possible because the coordinate frame of a chart may be composed with any constant invertible linear map (Definition 17.2), and the change of basis carrying the coordinate frame to \(\set{e_{i},f_{i}}\) is invertible. Let \(B_{0}\) be an open ball in the image of the chart, centred at the origin. Define on \(B_{0}\) the constant-coefficient two-form

all other components vanishing. Its matrix is exactly Equation (9.144), so by Step 1

and \(\dd\omega_{0}=0\) because its components are constants (Definition 17.103).

Step 3: the interpolation. Put \(\tau=\omega-\omega_{0}\) and

Each \(\omega_{t}\) is closed, since \(\omega\) and \(\omega_{0}\) are; and by Equation (A12.26), \(\left.\omega_{t}\right|_{P}=\left.\omega\right|_{P}\) for every \(t\), which is non-degenerate. The function

is continuous on \([0,1]\times B_{0}\) — a polynomial in the components, which are continuous — and \(D(t,0)\ne0\) for every \(t\). For each \(t\in[0,1]\) continuity therefore supplies an open interval \(I_{t}\ni t\) and a ball \(B^{(t)}\ni 0\) on which \(D\) does not vanish; the intervals \(I_{t}\) cover the compact \([0,1]\) (Theorem 10.11), finitely many of them suffice, and the intersection \(B_{1}\) of the corresponding finitely many balls is an open ball about the origin on which

Step 4: the primitive. \(B_{1}\) is star-shaped about the origin (Definition A12.10) and \(\tau\) is a closed two-form on it, so Lemma A12.11 gives a smooth one-form \(\sigma\) on \(B_{1}\) with

The vanishing of \(\sigma\) at the origin — which is Equation (A12.26) feeding through the explicit homotopy formula — is what will pin the point \(P\) and is the reason the argument is local rather than global.

Step 5: the vector field. Let \(W^{ij}(t,x)\) be the inverse of the matrix \(\left(\omega_{t}\right)_{ij}(x)\), which exists on \([0,1]\times B_{1}\) by Equation (A12.28) and whose entries are smooth there, being rational functions of the components with non-vanishing denominator \(D\). Define the time-dependent vector field

the equivalence because \(\left(i_{X}\omega_{t}\right)_{i}=X^{j}\left(\omega_{t}\right)_{ji}\), and contracting \(X^{j}(\omega_{t})_{ji}=-\sigma_{i}\) with \(W^{ik}\) returns Equation (A12.30). It is smooth in \((t,x)\), and by Equation (A12.29)

Step 6: the flow. By Equation (A12.31), Theorem A12.8 applies: there is a ball \(B_{2}\ni0\) and a smooth family \(\psi_{t}:B_{2}\longrightarrow B_{1}\), \(t\in[0,1]\), of diffeomorphisms onto open sets, with \(\psi_{0}=\id\), \(\pp_{t}\psi_{t}=X_{t}\circ\psi_{t}\) and \(\psi_{t}(0)=0\).

Step 7: the homotopy equation. By Lemma A12.9, Cartan's magic formula Equation (17.276), \(\dd\omega_{t}=0\), \(\pp_{t}\omega_{t}=\tau\) and Equation (A12.30),

where the last line used Equation (A12.29). Hence \(\psi_{t}^{\ast}\omega_{t}\) is independent of \(t\) on \(B_{2}\), and at \(t=0\) it equals \(\psi_{0}^{\ast}\omega_{0}=\omega_{0}\). Taking \(t=1\),

Step 8: reading off the chart. Let \(W=\psi_{1}(B_{2})\), an open neighbourhood of \(0\), and define on it

a chart of \(M\) around \(P\) with \(u(P)=0\), since \(\psi_{1}\) is a diffeomorphism and \(\psi_{1}(0)=0\). Applying \(\left(\psi_{1}^{-1}\right)^{\ast}\) to Equation (A12.33) and using Equation (A12.6), \(\omega = \left(\psi_{1}^{-1}\right)^{\ast}\omega_{0}\) on \(W\). By Equation (A12.7), \(\left(\psi_{1}^{-1}\right)^{\ast}\dd x^{a} = \dd\left(x^{a}\circ\psi_{1}^{-1}\right) = \dd u^{a}\), and by Equation (A12.4) the pullback distributes over the wedge, so Equation (A12.25) becomes

which is Equation (A12.1) with \(q^{i}=u^{i}\) and \(p_{i}=u^{m+i}\).