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Let $(M,\omega)$ be a symplectic manifold. Define a Lie bracket in $\Omega^1(M)$ requiring $\Bbb R$-bilinearity, skew-symmetry and $$[{\rm d}f,{\rm d}g] = {\rm d}\{f,g\}_\omega \quad\mbox{and}\quad [{\rm d}f,h\,{\rm d}g] = h[{\rm d}f,{\rm d}g] + X_f(h) {\rm d}g,$$where $X_f$ is the Hamiltonian vector field of $f$, satisfying $\omega(X_f,Y) = {\rm d}f(Y)$ for all $Y \in\mathfrak{X}(M)$, and the Poisson bracket is defined by $\{f,g\}_\omega = \omega(X_f,X_g)$.

I do not see how this defines the Lie bracket completely. Meaning that if we write in coordinates, say, $\alpha = \sum_i \alpha_i\,{\rm d}x^i$ and $\beta = \sum_{j}\beta_j\,{\rm d}x^j$, we'd have $$[\alpha,\beta] = \sum_{i,j}[\alpha_i\,{\rm d}x^i, \beta_j\,{\rm d}x^j],$$but I can't use the second condition to take out $\beta_j$, because $\alpha_i$ is still there. So I don't know how to get combinations of $[{\rm d}x^i, {\rm d}x^j]$, which are manageable using the first condition.

Can someone clarify this for me? Thanks.

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    $\begingroup$ There is a nice proof that the Jacobi identity holds, see here. $\endgroup$ May 9 '18 at 19:41
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    $\begingroup$ I suspect they were just too lazy to write out the complete Leibniz rule. $\endgroup$ May 9 '18 at 19:43
  • $\begingroup$ @Ted It is very possible. I'd guess the correct assumption is $$[h_1\,{\rm d}f, h_2\,{\rm d}g] = h_1h_2[{\rm d}f, {\rm d}g] + X_f(h_2){\rm d}g - X_g(h_1){\rm d}f,$$then? (mimicking what happens for vector fields) $\endgroup$
    – Ivo Terek
    May 9 '18 at 19:47
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    $\begingroup$ Oops. You're missing the $h_1$ on the second term and the $h_2$ on the last term. :) $\endgroup$ May 9 '18 at 19:48
  • $\begingroup$ Ok, thanks (Dietrich too, for the link)! I'll try to proceed with this, I think it'll work out fine. $\endgroup$
    – Ivo Terek
    May 9 '18 at 19:49
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Given the symplectic form $\omega$ on $M$, the relation $[\mathrm{d}f, \mathrm{d}g] = \mathrm{d}\{ f, g\}_{\omega}$ uniquely defines a Lie bracket on $\Omega^1(M)$, which is an avatar on 1-forms of the usual Lie bracket on vector fields (modulo a multiplicative sign which depends on conventions).

To be more precise, given $q \in M$, let $\omega^{\flat} : T_qM \to T_q^*M : v \mapsto v \lrcorner \, \omega_q$ be the interior product of a vector $v$ based at $q$ with $\omega_q$. Since $\omega_q$ is nondegenerate, this is an injection; since $T_qM$ and $T_q^*M$ have equal dimension $\mathrm{dim}(M)$, this is thus a bijection. This fiberwise bijection yields a (smooth) bundle isomorphism $\omega^{\flat} : TM \to T^*M$, which in turns yields a (smooth) isomorphism $\omega^{\flat} : \mathfrak{X}(M) \to \Omega^1(M)$ (as $C^{\infty}(M)$-modules).

What I meant in the first paragraph is then that we necessarily have (up to a sign) $$[\mathrm{d}f, \mathrm{d}g] = \omega^{\flat}[(\omega^{\flat})^{-1}(\mathrm{d}f), (\omega^{\flat})^{-1}(\mathrm{d}g)]$$ where the bracket on the right-hand side is the usual Lie bracket on vector fields.

This follows from well-known facts in symplectic geometry which allow to prove the equality $\mathrm{d}\{f,g\}_{\omega} = \omega^{\flat}[(\omega^{\flat})^{-1}(\mathrm{d}f), (\omega^{\flat})^{-1}(\mathrm{d}g)]$. First observe that the implicit definition of $X_f$ given by the equation $X_f \lrcorner \, \omega = \mathrm{d}f$ can be rewritten $X_f = (\omega^{\flat})^{-1}(\mathrm{d}f)$; thus we need to establish $$\mathrm{d}\{f,g\}_{\omega} = \omega^{\flat}[X_f, X_g] \; \mbox{ i.e. } \; \mathrm{d}\{f,g\}_{\omega} = [X_f, X_g] \lrcorner\, \omega,$$ a well-known expression. For the sake of completeness, let's prove it.

Lemma 1. $L_{X_f}\omega = 0$ (where $L_X$ denotes the Lie derivative with respect to the vector field $X$).

Proof. Using Cartan's formula, $$ L_{X_f}\omega = X_f \lrcorner \, \mathrm{d}\omega + \mathrm{d}(X_f \lrcorner \, \omega) = X_f \lrcorner \, \mathrm{d}\omega + \mathrm{d}\mathrm{d}f = X_f \lrcorner \, \mathrm{d}\omega. $$ Since $\omega$ is symplectic it is closed, hence the result. QED

Lemma 2. $\mathrm{d}\{f,g\}_{\omega} = [X_f, X_g] \lrcorner\, \omega$.

Proof. By definition, $\{f, g\}_{\omega} = \omega(X_f, X_g) = X_g \lrcorner \, (X_f \lrcorner \, \omega)$. Hence, using Cartan's formula and the Leibniz rule for the Lie derivative,

$$ \begin{align} \mathrm{d}\{f, g \}_{\omega} &= \mathrm{d}(X_g \lrcorner \, (X_f \lrcorner \, \omega)) = L_{X_g}(X_f \lrcorner \, \omega) - X_g\lrcorner \, \mathrm{d}(X_f \lrcorner \, \omega) \\ &= (L_{X_g}X_f) \lrcorner \, \omega + X_f \lrcorner \, (L_{X_g}\omega) - X_g \lrcorner \, \mathrm{d}\mathrm{d}f \\ &= (L_{X_g}X_f) \lrcorner \, \omega. \end{align}$$

Now, depending on conventions, we have $L_{X_g}X_f = \pm [X_f, X_g]$, say $'+'$. Hence the result. QED

That being established, the bracket on 1-forms is easily seen to be a Lie bracket, since it is 'conjugated' to the usual Lie bracket on vector fields. We can easily deduce for instance the Jacobi identity or the other 'defining' relation you gave in your post.

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