# The Jacobi Identity for the Poisson Bracket

It is well known that if $M, \Omega$ is a symplectic manifold then the Poisson bracket gives $C^\infty(M)$ the structure of a Lie algebra. The only way I have seen this proven is via a calculation in canonical coordinates, which I found rather unsatisfying. So I decided to try to prove it just by playing around with differential forms. I got quite far, but something isn't working out and I am hoping someone can help. Forgive me in advance for all the symbols.

Here is the setup. Given $f \in C^\infty(M)$, let $X_f$ denote the unique vector field which satisfies $\Omega(X_f, Y) = df(Y) = Y(f)$ for every vector field $Y$. We define the Poisson bracket of two functions $f$ and $g$ to be the smooth function $\{f, g \} = \Omega(X_f, X_g)$. I can show that the Poisson bracket is alternating and bilinear, but the Jacobi identity is giving me trouble. Here is what I have.

To start, let's try to get a handle on $\{ \{f, g \}, h\}$. Applying the definition, this is given by $d(\Omega(X_f, X_g))X_h$. So let's try to find an expression for $d(\Omega(X,Y))Z$ for arbitrary vector fields $X, Y, Z$.

Write $\Omega(X,Y) = i(Y)i(X)\Omega$ where $i(V)$ is the interior product by the vector field $V$. Applying Cartan's formula twice and using the fact that $\Omega$ is closed, we obtain the formula

$$d(\Omega(X,Y)) = (L_Y i(X) - i(Y) L_X) \Omega$$

where $L_V$ is the Lie derivative with respect to the vector field $V$. Using the identity $L_V i(W) - i(W) L_V = i([V,W])$, we get:

$$(L_Y i(X) - i(Y) L_X) = L_Y i(X) - L_X i(Y) + i([X,Y])$$

Now we plug in the vector field $Z$. We get $(L_Y i(X) \Omega)(Z) = Y(\Omega(X,Z)) - \Omega(X,[Y,Z])$ by the definition of the Lie derivative, and clearly $(i([X,Y])\Omega)(Z) = \Omega([X,Y],Z)$. Putting it all together:

$$d(\Omega(X,Y))Z = Y(\Omega(X,Z)) - X(\Omega(Y,Z)) + \Omega(Y, [X,Z]) - \Omega(X, [Y,Z]) + \Omega([X,Y], Z)$$

This simplifies dramatically in the case $X = X_f, Y = X_g, Z = X_h$. The difference of the first two terms simplifies to $$X_f, X_g$(h)$, and we get:

$$\{\{f, g\}, h\} = $X_f, X_g$(h) + $X_f, X_h$(g) - $X_g, X_h$(f) - $X_f, X_g$(h) = $X_f, X_h$(g) - $X_g, X_h$(f)$$

However, this final expression does not satisfy the Jacobi identity. It looks at first glance as though I just made a sign error somewhere; if the minus sign in the last expression were a plus sign, then the Jacobi identity would follow immediately. I have checked all of my signs as thoroughly as I can, and additionally I included all of my steps to demonstrate that if a different sign is inserted at any point in the argument then one obtains an equation in which the left hand side is alternating in two of its variables but the right hand side is not. Can anybody help?

-

The Jacobi identity for the Poisson bracket does indeed follow from the fact that $d\Omega =0$.

I claim that (twice) the Jacobi identity for functions $f,g,h$ is precisely $$d\Omega(X_f,X_g,X_h) = 0.$$

To see this, simply expand $d\Omega$.

You will find six terms of two kinds:

• three terms of the form $$X_f \Omega(X_g,X_h) = X_f \lbrace g,h \rbrace$$

• and three terms of the form $$\Omega([X_f,X_g],X_h).$$

To deal with the first kind of terms, notice that from the definition of $X_f$, for any function $g$, $$X_f g = \lbrace g, f \rbrace.$$ This means that $$X_f \Omega(X_g,X_h) = \lbrace \lbrace g,h \rbrace, f \rbrace.$$

To deal with the second kind of terms, notice that $$\iota_{[X_f,X_g]}\Omega = [L_{X_f},\iota_{X_g}]\Omega,$$ but since $d\Omega=0$, $$L_{X_f}\Omega = d \iota_{X_f}\Omega = 0,$$ and hence $$\iota_{[X_f,X_g]}\Omega = d \iota_{X_f}\iota_{X_g}\Omega = d\lbrace g,f\rbrace,$$ whence $$\Omega([X_f,X_g],X_h) = d\lbrace g,f\rbrace (X_h) = \lbrace\lbrace g,f\rbrace, h \rbrace.$$

Adding it all up you get twice the Jacobi identity.

-
I had to take care of some other stuff before I could come back to this and work through the details, but everything seemed to check out. Thanks! I wonder where the mistake lay in my original approach, which used more or less the same tools. –  Paul Siegel Jun 21 '10 at 20:47
There is no mistake in your calculations, as far as I can see. The last equation you write down is certainly correct. It simply requires extra work to derive the Jacobi identity from that point. I am afraid that the only way I can see to do this is basically to undo some of your identities and get back to something akin to what I wrote in my answer. –  José Figueroa-O'Farrill Jun 21 '10 at 23:00
An alternative after only proving $\iota_{[X_f,X_g]}\Omega=d\lbrace g,f\rbrace$ is to observe that the derivative of $\lbrace\lbrace f,g\rbrace h\rbrace+\lbrace\lbrace h,f\rbrace g\rbrace+\lbrace\lbrace g,h\rbrace f\rbrace$ is zero by the ordinary Jacobi identity. So it's locally constant. Since it's linear and locally defined, it must be zero (pick a point and kill it somewhere else with a bump function). –  Zack Apr 24 '11 at 7:23
@Josè Figueroa-O'Farrill: When, for an arbitrary almost-symplectic manifold, we again construct the bracket, is correct that $d\omega(X_f,X_g,X_h)$ is equal to the Jacobiator $J(f,g,h)$? or I am making same mistake? –  Giuseppe Tortorella Apr 24 '11 at 16:57
@Giuseppe: you are correct. Even if $\omega$ is not closed, if you define the hamiltonian vector field $X_f$ for a smooth function $f$ by $\omega(X_f,Y) = Yf$ for all vector fields $Y$, then again you will find that $d\omega(X_f,X_g,X_f)$ is (perhaps up to a sign) the "Jacobiator" of $f,g,h$. –  José Figueroa-O'Farrill Apr 26 '11 at 3:53
show 1 more comment

I wanted to add this as a comment to Jose's answer but it seems that I cannot do that as a new user.

For any bivector field $\sigma$, you can define a bracket on smooth functions by $\{f,g\} = \sigma(df, dg)$. This bracket is skew and will automatically satisfy the Liebniz rule. It will satisfy the Jacobi identity precisely when $[\sigma, \sigma] = 0$, where $[\cdot,\cdot]$ is the Schouten bracket. This point of view is important, for example, in defining Poisson cohomology.

Now suppose you take $\sigma = \omega^{-1}$, where $\omega$ is a nondegenerate 2-form. Jose's calcuation shows that $[\sigma, \sigma] = 0$ iff $d\omega = 0$.

-
I find your answer to be the right complement to Jose's answer. Thanks. –  Giuseppe Tortorella Apr 23 '11 at 21:37

Here's the way it's done in John Lee's Smooth Manifolds book:

$${\small \iota_{X_{\lbrace f,g\rbrace}}\omega = d\lbrace f,g \rbrace=d(X_gf) = d(\mathcal{L}_{X_g}f) =\mathcal{L}_{X_g}df=\mathcal{L}_{X_g}(\iota_{X_f}\omega)=\mathcal{\iota}_{[X_g,X_f]}\omega + \mathcal{L}_{X_g}\omega=\mathcal{\iota}_{[X_g,X_f]}\omega}$$ which by nondegeneracy of $\omega$ implies the desired result.

-
Oh woops. Didn't get to the punchline. I showed $$X_{\lbrace f,g\rbrace} = [X_g,X_f].$$ From here, do what Zack said. –  Dmitri Gekhtman Oct 18 '12 at 2:39

Perhaps one of the following references is helpful for you: a) page 12 of http://sundoc.bibliothek.uni-halle.de/habil-online/04/04A736/t3.pdf (unfortunately in German language), b) http://arxiv.org/PS_cache/physics/pdf/0210/0210074v1.pdf.

-

Let $(M,\omega)$ be a quasi-symplectic manifold.

Being $\omega$ a non-degenerate $2$-form on $M$, for any $f\in C^{\infty}(M)$ there exists a unique $X_f\in\mathcal{X}(M)$ such that $df=i(X_f)\omega$. The map $f\in C^{\infty}(M)\to X_f\in\mathcal{X}(M)$ is obviously $\mathbb{R}$-linear.

We introduce the pseudo-Poisson bracket over $C^{\infty}(M)$ by defining $\{f,g\}=X_f(g)\equiv \omega(X_g,X_f)$, for any $f$ and $g$ smooth function on $M$.
Immediately by definition $\{\cdot,\cdot\}:C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ is an antisymmetric $\mathbb{R}$-bilinear map.

We introduce also a sort of Jacobiator $J:C^{\infty}(M)\times C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$ by defining $J(f,g,h)=\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}$. This is a trilinear antisymmetric map which measure how much the Jacobi identity for the bracket is not satisfied.

I outline two different approaches.

1)$d\omega=0$ implies the Jacobi identity for $\{\cdot,\cdot\}$.

The Jacobi identity for the pseudo-Poisson bracket can be easily rewritten as (*)$X_{\{f,g\}}=[X_f,X_g]$, for any $f$ and $g$ smooth functions on $M$. So $C^{\infty}(M),\{\cdot,\cdot\})$ is a Lie algebra if and only if the map $(C^{\infty}(M),\{\cdot,\cdot\})\ni f\to X_f\in(\mathcal{X}(M),[\cdot,\cdot])$ is a homomorphism of $\mathbb{R}$-algebras.

Being $d\omega=0$, by the H.Cartan's formula we get that a smooth vector field $X$ on $(M,\omega)$ is symplectic, i.e. $\mathcal{L}(X)(\omega)=0$, if and only if it is locally hamiltonian, i.e. $d.i(X)\omega=0$.

Now the condition (*) is a consequence of the much more strong statement:
Theorem.If $Y$ and $Z$ are symplectic vector fields on $(M,\omega)$, i.e. $\mathcal{L}(Y)(\omega)=\mathcal{L}(Z)(\omega)=0$, then $[Y,Z]=-X_{\omega(Y,Z)}$, i.e. $[Y,Z]$ is a hamiltonian vector field with $-\omega(Y,Z)$ as Hamilton function.
Proof. $i([Y,Z])\omega=\mathcal{L}(Y).i(Z)\omega-i(Z).\mathcal{L}(Y)\omega=d.i(Y).i(Z)\omega+i(Y).d.i(Z)\omega=d(\omega(Z,Y))$.
(Having used the hypothesis, the H.Cartan's formula, and the formula $[\mathcal{L}(Y),i(Z)]=i([Y,Z])$).

2)$(d\omega)(X_f,X_g,X_h)=J(f,g,h)$, for any $f$,$g$,and $h$ smooth functions on $M$.

By Palais' expression of exterior derivative through Lie derivatives, we can express $(d\omega)(X_f,X_g,X_h)$ as the sum of two terms obtained respectively by $\mathcal{L}(X_f)(\omega(X_g,X_h)$ and by $\omega(X_f,[X_g,X_h])$ summing over the ciclic permutations of $(f,g,h)$.

Now $\mathcal{L}(X_f)(\omega(X_g,X_h)=-\{f,\{g,h\}\}$ and $\omega(X_f,[X_g,X_h])=\{g,\{h,f\}\}+\{h,\{f,g\}\}$, and so we get the thesis.

-