Say we have a constraint system with second-class constraints $\chi_N(q,p)=0$. To define Dirac brackets we need the Poisson brackets of these constraints: $C_{NM}=\{\chi_N(q,p),\chi_M(q,p)\}_P$ . Is the matrix $C_{NM}$ independent of $q,p$?

The motivation of the question is that when I tried to verify Dirac brackets satisfy Jacobi identity, it seems I need to assume $C_{NM}$ doesn't depend on canonical coordinates(To be honest, I'm not 100% sure if this assumption is necessary, because I did not try very hard on the verification without this assumption because one side of the Jacobi identity will become horribly complicated). Besides, for some simple examples(e.g. a massive vector field), it does turn out to be the case.

EDIT: I think I know where I made mistakes, somewhere in the calculation I needed some terms to be 0, I was searching in my brain only all the strong equalities, while in fact some weak equalities could help.

  • $\begingroup$ Thanks for the replies from twistor59 and @Qmechanic, but recently I haven't got the time to grab a pen and a piece of paper to verify what you said. I should be able to think about it in more detail in a few days when I'm free. $\endgroup$
    – Jia Yiyang
    Mar 5, 2013 at 10:55

2 Answers 2


Your constraint matrix $C_{NM} = \{\chi_{N}, \chi_{M}\}_{P} \ $ is defined on the whole of phase space and is invertible everywhere, but there's no reason for it to be constant.

Perhaps the easiest way to see that the Dirac bracket obeys the Jacobi identity is to note that the Jacobi identity for the case of Poisson brackets holds because the symplectic form is closed. If $M$ is phase space and we look at the submanifold $M_s \subset M$ selected by imposing the second class constaints, then the form defining the Dirac bracket is the pullback of the symplectic form via the embedding $M_s \hookrightarrow M$ and hence is closed too.

  • $\begingroup$ Interesting, but is it hard or tedious to prove it is the pullback of the embedding? $\endgroup$
    – Jia Yiyang
    Mar 11, 2013 at 13:08

Given an $2n$-dimensional symplectic manifold $(M,\omega)$ with corresponding Poisson bracket $$\{\cdot,\cdot\}_{PB}~:~ C^{\infty}(M)\times C^{\infty}(M) \to C^{\infty}(M),$$

and physical submanifold

$$\Sigma~:=~\{ z\in M ~|~ \chi^{1}(z)~=~0, \ldots,\chi^{2m}(z)~=~0 \}~\subseteq~ M$$

defined by a set of $2m$ (for simplicity globally defined) second-class constraints $\chi^1$, $\ldots$, $\chi^{2m}$. Here $m\leq n$. The second-class condition is equivalent to that the matrix

$$\Delta^{ab}~:=~\{ \chi^{a} ,\chi^{b}\}_{PB} $$

restricted to (and hence in some open neighborhood of) the physical subspace $\Sigma$ is an invertible matrix. To answer OP's question, the matrix $\Delta^{ab}$ depends in general on the point $z\in M$ and, in particular, it does not need to be constant. (Let us here for simplicity assume that $\Delta^{ab}$ is an invertible matrix globally on the whole $M$.)

The Dirac bracket

$$\{\cdot,\cdot\}_{DB}~:~ C^{\infty}(M)\times C^{\infty}(M) \to C^{\infty}(M),$$

is defined as

$$ \{f,g\}_{DB} ~:=~ \{f, g\}_{PB}-\{f, \chi^{a}\}_{PB} (\Delta^{-1})_{ab}\{ \chi^{b},g\}_{PB}. $$

The Dirac bracket is a non-invertible Poisson structure of rank $2(n-m)$ on the full space $M$.

Concerning the Jacobi identity for the Dirac bracket

$$ \{\{f,g\}_{DB},h\}_{DB} +\text{cycl.}(f,g,h) ~=~0, $$

it is a rather remarkable fact that the Jacobi identity holds strongly in the full space $M$ without imposing the second-class constraints. For all physical purposes, it would have been enough if the Jacobi identity only holds weakly modulo second-class constraints, but remarkably the Dirac construction yields a strong Jacobi identity on the full space $M$.

Moreover, OP might find it interesting that Dirac himself wrote in Ref. 1

[...]I don't know of any neat way of proving the Jacobi identity for the [Dirac bracket]. If one just substitutes according to the definition and works it out in a complicated way, one does find that all the terms cancel out and the left-hand side equals zero. [...]


  1. P.A.M. Dirac, Lectures on Quantum Mechanics, (1964) p.42.
  • $\begingroup$ I tied the brute-force calculation today(without assuming the matrix to be independent of canonical coordinates), but failed to cancel the additional terms, do you have any reference where an explicit calculation is presented? The quote from Dirac is also interesting, but then I would wonder how he came up with his bracket. Was it like Dirac invented this due to some other considerations and Jacobi identity just emerged for free? $\endgroup$
    – Jia Yiyang
    Mar 11, 2013 at 13:12

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