Editing Poisson bracket (section)

==The Poisson bracket in coordinate-free language==
Let <math>M</math> be a [[symplectic manifold]], that is, a [[manifold]] equipped with a [[symplectic form]]: a [[Differential form|2-form]] <math>\omega</math> which is both '''closed''' (i.e., its [[exterior derivative]] <math>d \omega</math> vanishes) and '''non-degenerate'''.  For example, in the treatment above, take <math>M</math> to be <math>\mathbb{R}^{2n}</math> and take
<math display="block">\omega = \sum_{i=1}^{n} d q_i \wedge d p_i.</math>

If <math> \iota_v \omega</math> is the [[interior product]] or [[Tensor contraction|contraction]] operation defined by <math> (\iota_v \omega)(u) =  \omega(v,\, u)</math>, then non-degeneracy is equivalent to saying that for every one-form <math>\alpha</math> there is a unique vector field <math>\Omega_\alpha</math> such that <math> \iota_{\Omega_\alpha} \omega =  \alpha</math>. Alternatively, <math> \Omega_{d H} = \omega^{-1}(d H)</math>. Then if <math>H</math> is a smooth function on <math>M</math>, the [[Hamiltonian vector field]] <math>X_H</math> can be defined to be <math> \Omega_{d H}</math>.  It is easy to see that
<math display="block">\begin{align}
  X_{p_i} &=  \frac{\partial}{\partial q_i} \\
  X_{q_i} &= -\frac{\partial}{\partial p_i}.
\end{align}</math>

The '''Poisson bracket''' <math>\ \{\cdot,\, \cdot\} </math> on {{math|(''M'', ''ω'')}} is a [[bilinear map|bilinear operation]] on [[differentiable function]]s, defined by <math> \{f,\, g\} \;=\; \omega(X_f,\, X_g) </math>; the Poisson bracket of two functions on {{math|''M''}} is itself a function on {{math|''M''}}.  The Poisson bracket is antisymmetric because:
<math display="block">\{f, g\} = \omega(X_f, X_g) = -\omega(X_g, X_f) = -\{g, f\} .</math>

Furthermore,
{{NumBlk||<math display="block">\begin{align}
  \{f, g\} &= \omega(X_f, X_g) = \omega(\Omega_{df}, X_g) \\
           &= (\iota_{\Omega_{df}}\omega)(X_g) = df(X_g) \\
           &= X_g f = \mathcal{L}_{X_g} f.
\end{align}</math>|{{EquationRef|1}}}}

Here {{math|''X<sub>g</sub>f''}} denotes the vector field {{math|''X<sub>g</sub>''}} applied to the function {{math|''f''}} as a directional derivative, and <math>\mathcal{L}_{X_g} f</math> denotes the (entirely equivalent) [[Lie derivative]] of the function {{math|''f''}}.

If {{math|α}} is an arbitrary one-form on {{math|''M''}}, the vector field {{math|Ω<sub>α</sub>}} generates (at least locally) a [[flow (mathematics)|flow]] <math> \phi_x(t)</math> satisfying the boundary condition <math> \phi_x(0) = x</math> and the first-order differential equation
<math display="block">\frac{d\phi_x}{dt} = \left. \Omega_\alpha \right|_{\phi_x(t)}.</math>

The <math> \phi_x(t)</math> will be [[symplectomorphism]]s ([[canonical transformation]]s) for every {{math|''t''}} as a function of {{math|''x''}} if and only if <math> \mathcal{L}_{\Omega_\alpha}\omega \;=\; 0</math>; when this is true, {{math|Ω<sub>α</sub>}} is called a [[symplectic vector field]].  Recalling [[Cartan's identity]] <math> \mathcal{L}_X\omega \;=\; d (\iota_X \omega) \,+\, \iota_X d\omega</math> and {{math|1=''d''ω = 0}}, it follows that <math> \mathcal{L}_{\Omega_\alpha}\omega \;=\; d\left(\iota_{\Omega_\alpha} \omega\right) \;=\; d\alpha</math>. Therefore, {{math|Ω<sub>α</sub>}} is a symplectic vector field if and only if α is a [[Closed and exact differential forms|closed form]].  Since <math> d(df) \;=\; d^2f \;=\; 0</math>, it follows that every Hamiltonian vector field {{math|''X<sub>f</sub>''}} is a symplectic vector field, and that the Hamiltonian flow consists of canonical transformations.  From {{EquationNote|1|(1)}} above, under the Hamiltonian flow <math>X_\mathcal H</math>,
<math display="block">\frac{d}{dt}f(\phi_x(t)) = X_\mathcal{H}f = \{f,\mathcal H\}.</math>

This is a fundamental result in Hamiltonian mechanics, governing the time evolution of functions defined on phase space.  As noted above, when {{math|1={''f'',''\mathcal H''} = 0}}, {{math|''f''}} is a constant of motion of the system.  In addition, in canonical coordinates (with <math> \{p_i,\, p_j\} \;=\; \{q_i,q_j\} \;=\; 0</math> and <math>\{q_i,\, p_j\} \;=\; \delta_{ij}</math>), Hamilton's equations for the time evolution of the system follow immediately from this formula.

It also follows from {{EquationNote|1|(1)}} that the Poisson bracket is a [[derivation (abstract algebra)|derivation]]; that is, it satisfies a non-commutative version of Leibniz's [[product rule]]:
{{NumBlk||<math display="block">\{fg,h\} = f\{g,h\} + g\{f,h\},</math> and <math display="block">\{f,gh\} = g\{f,h\} + h\{f,g\}.</math>|{{EquationRef|2}}}}

The Poisson bracket is intimately connected to the [[Lie bracket of vector fields|Lie bracket]] of the Hamiltonian vector fields.  Because the Lie derivative is a derivation,
<math display="block">\mathcal L_v\iota_u\omega = \iota_{\mathcal L_vu}\omega + \iota_u\mathcal L_v\omega = \iota_{[v,u]}\omega + \iota_u\mathcal L_v\omega.</math>

Thus if {{math|''v''}} and {{math|''u''}} are symplectic, using <math> \mathcal{L}_v\omega =0=\mathcal L_u\omega</math>, Cartan's identity, and the fact that <math>\iota_u\omega</math> is a closed form,
<math display="block">\iota_{[v,u]}\omega = \mathcal L_v\iota_u\omega = d(\iota_v\iota_u\omega) + \iota_vd(\iota_u\omega) = d(\iota_v\iota_u\omega) = d(\omega(u,v)).</math>

It follows that <math>[v,u] = X_{\omega(u,v)}</math>, so that
{{NumBlk||<math display="block">[X_f,X_g] = X_{\omega(X_g,X_f)} = -X_{\omega(X_f,X_g)} = -X_{\{f,g\}}.</math>|{{EquationRef|3}}}}

Thus, the Poisson bracket on functions corresponds to the Lie bracket of the associated Hamiltonian vector fields.  We have also shown that the Lie bracket of two symplectic vector fields is a Hamiltonian vector field and hence is also symplectic.  In the language of [[abstract algebra]], the symplectic vector fields form a [[subalgebra]] of the [[Lie algebra]] of smooth vector fields on {{math|''M''}}, and the Hamiltonian vector fields form an [[algebraic ideal|ideal]] of this subalgebra.  The symplectic vector fields are the Lie algebra of the (infinite-dimensional) [[Lie group]] of [[symplectomorphism]]s of {{math|''M''}}.

It is widely asserted that the [[Jacobi identity]] for the Poisson bracket,
<math display="block">\{f,\{g,h\}\} + \{g,\{h,f\}\} + \{h,\{f,g\}\} = 0</math>
follows from the corresponding identity for the Lie bracket of vector fields, but this is true only up to a locally constant function.  However, to prove the Jacobi identity for the Poisson bracket, it is [[Jacobi identity#Examples|sufficient]] to show that:
<math display="block">\operatorname{ad}_{\{g,f\}}=\operatorname{ad}_{-\{f,g\}}=[\operatorname{ad}_f,\operatorname{ad}_g]</math>
where the operator <math>\operatorname{ad}_g</math> on smooth functions on {{math|''M''}} is defined by <math>\operatorname{ad}_g(\cdot) \;=\; \{\cdot,\, g\}</math> and the bracket on the right-hand side is the commutator of operators, <math> [\operatorname A,\, \operatorname B] \;=\; \operatorname A\operatorname B - \operatorname B\operatorname A</math>.  By {{EquationNote|1|(1)}}, the operator <math>\operatorname{ad}_g</math> is equal to the operator {{math|''X<sub>g</sub>''}}.  The proof of the Jacobi identity follows from {{EquationNote|3|(3)}} because, up to the factor of -1, the Lie bracket of vector fields is just their commutator as differential operators.

The [[Algebra over a field|algebra]] of smooth functions on M, together with the Poisson bracket forms a [[Poisson algebra]], because it is a [[Lie algebra]] under the Poisson bracket, which additionally satisfies Leibniz's rule {{EquationNote|2|(2)}}.  We have shown that every [[symplectic manifold]] is a [[Poisson manifold]], that is a manifold with a "curly-bracket" operator on smooth functions such that the smooth functions form a Poisson algebra.  However, not every Poisson manifold arises in this way, because Poisson manifolds allow for degeneracy which cannot arise in the symplectic case.