# Poisson brackets for Hamiltonian descriptions

1. Feb 5, 2009

### haushofer

Hi, I have a (maybe rather technical) question about the Hamiltonian formulation of gauge theories, which I don't get.

With an infinitesimal symmetry on your space-time M one can look at the corresponding transformation of the canonical variables in phase-space PS. This can be done by a phase space function H[]:M --> PS. The generators of these symmetries give you global charges.

If I generate my symmetry with a vector field $\xi$ or $\alpha$, the statement which is often made is that

$$\{H[\xi],H[\alpha] \} = H [[\xi,\alpha]]$$

In words: the Poisson algebra {} of the charges (LHS) is isomorphic (=) to the Lie algebra of the corresponding symmetries on your space-time (RHS).

My question is: why is this so natural to assume? Henneaux and Brown wrote an article about the details and adjustments of this assumption (central charges in the canonical realization...), but I don't see why this should be "natural" in the first place. I'm missing something here.

Thanks in forward :)

2. Feb 7, 2009

### samalkhaiat

I hope, I have uderstood your question correctly!

In the real life, we use Noether procedure to write $H_{\alpha}$ in terms of the dynamical variables on PS:

$$H_{\alpha} = \int dx \pi \mathcal{L}_{\alpha} \phi \ \ \ (R)$$

where $\mathcal{L}_{\alpha}$ is Lie derivative along the vector field $\alpha$.
It follows immediately from eq(R) that

1) canonical transformations generated by the charge/constraint correspond precisely to the diffeomorphisms generated by the vector field $\alpha$ ,i.e.,

$$\{ H_{\alpha} , \phi \} = \mathcal{L}_{\alpha} \phi, \ \ \mbox{etc.} \ (1)$$

2) charges/constraints satisfy the algebra

$$\{ H_{\alpha}, H_{\xi} \} = H_{[\alpha , \xi ]} \ \ \ (2)$$

To confirm the consistency between eq{1) and eq(2), we use the Jacobi identity

$$\{ \{ H_{\alpha} , H_{\xi} \} , \phi \} = \{\{H_{\alpha}, \phi\} , H_{\xi} \} - \{\{H_{\xi} , \phi \} , H_{\alpha} \}$$

This gives

$$\{\{H_{\alpha} , H_{\xi} \} , \phi \} = \mathcal{L}_{\alpha} \mathcal{L}_{\xi} \phi - \mathcal{L}_{\xi}\mathcal{L}_{\alpha} \phi = \mathcal{L}_{[\alpha , \xi ]} \phi$$

However, the algebra of constraints is not, in general, a proper Lie algebra: the RHS of eq(2) may contain structure functions rather than structure constants. This means, in the BRST terminology, that we may have an open algebra. Indeed, it has been shown [Bargmann and Komar] that the algebra of constraints is not isomorphic to the Lie algebra of the "gauge group" of tetrad-gravity.

*****

Ok, here is a math question for you:

Given two function $f,g : PS \rightarrow \mathbb{R}$, their Poisson bracket is defined by

$$\{ f , g \} = \mathcal{L}_{X_{f}}g = - \mathcal{L}_{X_{g}}f$$

this means that PB turns the vector space of functions on PS into Lie algebra. Use this Lie-bracket to show that the linear map $f \rightarrow X_{f}$ (which associates to f its (Hamiltonian) vector field) takes Poisson brackets of functions into commutators of vector fields:

$$[X_{f} , X_{g}] = X_{\{f,g\}} \ \ (3)$$

compair this with eq(2).

regards

sam

3. Feb 9, 2009

### haushofer

A, nice! I will look at this tomorrow! Sam to the rescue, again :D

4. Feb 11, 2009

### samalkhaiat

It is a pleasure, very time.