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A first-class constraint generates an infinitesimal canonical transformation $\delta F=\{F,{ϵ}^i{\phi }_i\}$. For the phase-space action $I=\int (p\dot{q}-{\lambda }^i{\phi }_i)dt$, the multiplier transformation $\delta {\lambda }^k={\dot{ϵ}}^k-{\lambda }^i{ϵ}^j{C}_{ij}{}^k$ cancels the bulk variation when $\{{\phi }_i,{\phi }_j\}=C_{ij}{}^k{\phi }_k$. The remaining temporal boundary term vanishes for parameters that vanish at the endpoints.

The Poisson bracket in the original coordinates isFor $z=(q,p)$, setThen ${\omega }_0(u,v)=u^{T}Jv$ and $\{F,G\}=\nabla F^{T}J\nabla G$. The pullback definition of a canonical transformation gives the pointwise matrix conditionSince $A$ is invertible, this is equivalent to $AJ{A}^T=J$. Indeed, invert $A^{T}JA=J$ and use $J^{-1}=-J$ to obtain $A^{-1}J{A}^{-T}=J$, then multiply by $A$ and $A^T$. Applying the same calculation to the inverse implication gives equivalence. These are the two equivalent symplectic matrix identities that connect the symplectic form with Poisson brackets.

The chain rule yieldsIf $f$ is a canonical transformation, the symplectic matrix identity $AJ{A}^T=J$ makes this expression $\{F,G{\}}_{Q,P}\circ f$. Thus preservation of the symplectic form implies preservation of every Poisson bracket, for all $C^1$ functions.

Conversely, assume this Poisson bracket identity for every $F,G$. Take $F(Z)=u\cdot Z$ and $G(Z)=v\cdot Z$, for arbitrary constant vectors $u,v\in {\mathbb{R}}^{2n}$. It gives $u^{T}AJ{A}^{T}v=u^{T}Jv$. Since this holds for every $u,v$, $AJ{A}^T=J$, hence $A^{T}JA=J$. Therefore preservation of all Poisson brackets is equivalent to canonicity. Only first derivatives of $f,F,G$ enter this argument.

Apply preservation of Poisson brackets to the coordinate functions. It givesConversely, the entries of $AJ{A}^T$ are exactly the Poisson brackets between all transformed coordinate functions. The displayed relations therefore say $AJ{A}^T=J$, which implies the canonical transformation condition and, by the chain rule, preservation of every Poisson bracket. Thus all three conditions are equivalent:In particular the original coordinate bracket $\{q_i,p_j\}$ is ${\delta }_{ij}$ with the chosen Poisson bracket convention.

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 Poincare 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.

Write $D={\partial }_{\mu }{A}^{\mu }$. The additional gauge fixing variation is $\delta (-D^2/2)=-D{\partial }_{\mu }\delta A^{\mu }$. Integration by parts gives the Euler-Lagrange field equation ${\partial }_{\mu }{F}^{\mu \nu }+{\partial }^{\nu }D=0$, hence $\square A^{\nu }=0$. This equation follows from the gauge-fixed density without imposing $D=0$ as a separate operator identity.

Direct differentiation of the density as printed gives the canonical momentum components conjugate to the lower-index fields:There is a boundary-term convention to reconcile with part (c). The Feynman-gauge Maxwell kinetic density after a boundary-term subtraction isIt has the same Euler-Lagrange field equations, but its canonical momentum components are ${\pi }^{\prime \nu }=-{\dot{A}}^{\nu }$. The mode expansion supplied in part (c) is the expansion of these latter momenta. Thus it is valid after the stated boundary-term subtraction; it is not the direct derivative of the original density. For example, a time-independent spatially varying $A_0$ with $A_i=0$ gives ${\pi }^i=-{\partial }_i{A}_0$ but ${\pi }^{\prime i}=0$.

The boundary-term shift of canonical field momenta is a canonical transformation. If spatial boundary terms vanish, $B=\int d^{3}x{K}^0=\int d^{3}x{A}_0{\partial }_i{A}_i$, and ${\pi }^{\nu }={\pi }^{\prime \nu }+\delta B/\delta A_{\nu }$. This explains why the two conventions have the same dynamics while their momentum formulas differ.