# Connection between Poisson Brackets and Symplectic Form

Jose and Saletan say the matrix elements of the Poisson Brackets (PB) in the ${q,p}$ basis are the same as those of the inverse of the symplectic matrix $\Omega^{-1}$, whereas the matrix elements of the symplectic form are equal to those of the symplectic matrix $\Omega$ .

I have no problem with the latter statement, but I do with the first. That's because the PBs are introducing by writing Hamilton's equation as

$\dot{\xi^j} = \omega^{jk}{\frac{\partial H }{\partial \xi^k}}$,

where $\omega^{jk}$ are the elements of the $\Omega$, and then taking the Lie derivative of a dynamical variable $f$ along the dynamical vector field, which gives $L_{\Delta}f = (\partial_j f) \omega^{jk}(\partial_k H)+ \partial_{t}f$. It is later said that the term containing the $\omega^{jk}$ is the PB $\left \{ f,H \right \}$, which I have no problem at all, as it gives the right expression for the PBs of the canonical coordinates when the $\omega^{jk}$ are the elements of the symplectic matrix $\Omega$, i.e., in the way it was first introduced through Hamilton's equations. However, as I have mentioned, in a later consideration they say that that the $\omega^{jk}$ of the PBs are the elements of $\Omega^{-1}$, which made me confused, especially because at several times in the book they use $\omega^{jk}$ as components of $\Omega$, and $\Omega_{jk}$ as components of $\Omega^{-1}$, at various derivations. However, I do not believe the statement on the book about the elements of the PBs being those of $\Omega^{-1}$ is wrong, because this is used in the derivation of the preservation of the symplectic form under canonical transformations. Therefore, I think there is a misconception on my behalf somewhere, which I do not know where it is, and I would be thankful if anyone could shed some light on this.

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## 4 Answers

It seems that OP just wants to trace a sign convention in a specific book (Ref. [JS]). To answer this most convincingly, we should document where we read what.

• In the case of canonical coordinates (also known as Darboux coordinates), [JS] uses a convention where the positions $q^i$ are ordered before the momenta $p_i$, $$\xi^i~=~(q^1, \ldots, q^n,p_1, \ldots p_n),$$ cf. e.g. p. 215 in [JS].

• On top of p. 230 in [JS] is written:

The elements $\omega_{jk}$ of the symplectic form and the $\omega^{jk}$ of the Poisson bracket [...] are the matrix elements of $\Omega$ and $\Omega^{-1}$, respectively.

Since we are only after a sign convention, let us for simplicity restrict to canonical/Darboux coordinates. Then $\Omega^2=-1_{2n}$, and hence the difference between $\Omega$ and $\Omega^{-1}$ boils down to a sign.

• On p. 216 eqs. (5.42b) and (5.43) read $$\tag{5.42b} \omega_{\ell j} \dot{\xi}^j~=~ \partial_{\ell} H,$$ $$\tag{5.43} \Omega ~=~\begin{bmatrix} 0_n & -I_n \cr I_n & 0_n \end{bmatrix}.$$

• On p. 218 eq. (5.47) defines the Poisson bracket $$\tag{5.47} \{f,g\} ~\equiv~ (\partial_jf) \omega^{jk} (\partial_kg) ~\equiv~ \frac{\partial f}{\partial q^{\alpha}}\frac{\partial g}{\partial p_{\alpha}}-\frac{\partial f}{\partial p_{\alpha}}\frac{\partial g}{\partial q^{\alpha}},$$ leading to Hamilton's equation of motion (5.50), $$\tag{5.50} \dot{\xi}^j~=~ \{\xi^j , H\}.$$

We conclude that [JS] has the convention that

$$\Omega^{-1} ~=~\begin{bmatrix} 0_n & I_n \cr -I_n & 0_n \end{bmatrix}.$$

So far so good.

• On p.228 is written

$$\tag{5.75} \omega ~=~ dq^{\alpha} \wedge dp_{\alpha}.$$

Comparing with $\Omega$ and $\omega_{ij}$, we conclude that [JS] has the convention that

$$\omega ~=~ \frac{1}{2} \omega_{ij} d\xi^j \wedge d\xi^i ~=~ -\frac{1}{2} \omega_{ij} d\xi^i \wedge d\xi^j .$$

Note the opposite ordering of $i$ and $j$! This is probably the point where OP and many others instead would have liked to defined it oppositely as

$$\tag{Opposite JS} \omega ~=~ \frac{1}{2} \omega_{ij} d\xi^i \wedge d\xi^j,$$

and

$$\tag{Opposite JS} \omega ~=~ dp_{\alpha} \wedge dq^{\alpha}.$$

References:

[JS] Jose and Saletan, Classical Dynamics: A contemporary Approach, 1998.

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Oh, so they did do an index swap. This is a very wrong convention--- it conflicts with what you would do with a metric manifold. They should have defined one of the two with a minus sign. OP was right to get confused. Thanks for tracking it down, +1. – Ron Maimon May 14 '12 at 1:42
Thanks for the clarification. It still doesn't make any sense to me why they chose such confusing way to present this, but at least I understand what happened now. I do have to say though that in my version of the book (also from 1998) the matrix you have in eq. 5.43 is for $\Omega ^{-1}$ not $\Omega$, and as a result your matrix after "we conclude..." should be $\Omega$. It doesn't change anything else you pointed out, but could create confusion for someone else. – Raphael R. Jun 11 '12 at 9:29
I have now checked again my 1998 version of [JS]. Its eq. (5.43) is for $\Omega$ not $\Omega^{-1}$. – Qmechanic♦ Jun 28 '12 at 21:06

Let's see: A symplectic form on a manifold $P$ (the phasespace) is a nondegenerate closed two-form $\omega$.

This gives you for every function $f \colon P \to \mathbf{R}$ a vectorfield $\xi_f$ defined by

$$i_{\xi_f}\omega = -df$$,

where $i$ denotes the interior product.

Then for two functions $f,g \colon P \to \mathbf{R}$ the Poissonbracket is

$$\{f,g\} = \xi_f g = \omega(\xi_f,\xi_g) = - \{g,f\}$$

Now in local coordinates $\xi_f = \xi^i_f \partial_i$, so you get

$$i_{\xi_f}\omega(\partial_j) = \omega(\xi^i_f \partial_i,\partial_j) = \xi^i_f \omega(\partial_i,\partial_j) = \xi^i_f \omega_{ij} = -df(\partial_j) = -\partial_j f$$

Since $\omega$ is invertible this means $\xi^i_f = -\omega^{ij}\partial_j f$, hence $$\{f,g\} = \omega^{ij}\partial_i f \partial_j g$$

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I do not see how you answered the confusion exposed on my post. I know the matrices representing the PB and the symplectic form are inverse to one another, the problem lies on their representation, i.e., in the (q,p) representation w^{ij} have to be the matrix elements of the symplectic matrix so we get the right expression for the PB (if the order of the \xi are q1,q2,p1,p2, for example). Whereas, if I use w_(ij) as the matrix elements of the inverse of omega to obtain the symplectic form I get (again in the q,p rep), w = - dq ^ dp instead of w = dq ^ dp. – Raphael R. Mar 10 '12 at 5:05

I don't have Jose and Saletan, but see Theorem 18.1.3, in particular (18.6), in my book Classical and Quantum Mechanics via Lie algebras

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The issue here is the raising and lowering of indices. The form $\omega_{ij}$ with lower indices is not the same as the two-tensor $\omega^{ij}$, although with a flat metric (as your examples have), the two are the same (since raising and lowering an index makes no difference). But there is a sign issue--- the two tensors are antisymmetric, and you could define raising the index to swap the index position too. In this case, you get an extra minus sign.

The issue is important, because the symplectic matrix has one down and one up index, and squares to -1:

$$\omega^i_j \omega ^j_k = -\delta^i_k$$

(assuming no swap definition). Then lowering and raising with the metric,

$$\omega^{ij} = g^{jk}\omega^i_k$$

$$\omega_{ij} = g_{ik}\omega^k_i$$

so that if you multiply the entries as matrices,

$$\omega^{ij}\omega_{jk} = \omega^i_k g^{kj}g_{jl}\omega^l_k = -\delta^i_k$$

since the g's are inverse to each other, so you get cancellation in the middle. The result is that the entries of the upper index $\omega$ are the inverse matrix (up to a sign) of the lower index $\omega$, and this is what the authors are trying to say.

They are either sloppy about the sign, or have an index flipping convention that fixes it up, I don't know. But the sign on the inverse is the reason for your confusion. I wouldn't use their terminology--- I would say that the upper index $\omega$ has entries which are the negative inverse of the lower index $\omega$'s, but they probably fix up any signs using their experience and intuition, so that the formulas end up correct in the end.

### EDIT: Qmechanic found the reference

It's a swap convention. They flip the indices for the tensor vs. the form, absorbing the minus sign. This is not a great convention, but it's what they do. Thanks Qmechanic for figuring it out.

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Comment to the answer(v1): The notions of a symplectic 2-form $\omega_{ij}$ and a Poisson bi-vector field  $\omega^{ij}$ can (and should) be defined without relying on the existence of an arbitrary choice of a metric tensor field $g_{ij}$. – Qmechanic♦ May 11 '12 at 10:20
@RonMaimon: bedroom, haha. – Nick Kidman May 11 '12 at 14:46
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