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:


$$
\begin{align}
\{\{f, g\}, h\} &= [ X\_f, X\_g ](h) + [ X\_f, X\_h ](g) - [ X\_g, X\_h ](f) - [ X\_f, X\_g ](h)\\
&= [ X\_f, X\_h ](g) - [ X\_g, X\_h ](f)
\end{align}$$


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
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.


