# Proof of Poincaré algebra with Poisson bracket

I'm trying to prove that the generators of Poincaré group in Poisson bracket close the well-known Poincaré algebra. In particular, I don't know how to prove that $$\{P_\mu,P_\nu\}=0$$.

Let's take an infinitesimal translation in space-time, such that $$x^\mu\rightarrow x'^\mu=x^\mu+\epsilon^\mu\Rightarrow\delta x^\mu=\epsilon^\mu \\\phi(x)\rightarrow\phi'(x)=\phi(x-\epsilon)=\phi(x)-\epsilon^\mu\partial_\mu\phi(x) \Rightarrow \delta_0\phi(x)=-\epsilon^\mu\partial_\mu\phi(x)$$ and suppose that our physical system has an action which is invariant under this translation $$\delta S|_{off-shell}=0$$ From Noether's theorem we know that we have a conserved charge, so in this case we have 4, one for each direction. And we also know that these charges are the components of the four-momentum

$$P^\mu=\int_\sigma d\sigma_\nu T^{\nu\mu}(x)$$

where $$d\sigma_\nu=n_\nu d\sigma$$ ($$n_\nu$$ is the vector normal to the space-like surface $$\sigma$$) and $$T^{\mu\nu}=\pi^\mu\phi^\nu-\mathcal Lg^{\mu\nu}$$ is the stress-energy tensor ($$\pi^\mu=\frac{\partial \mathcal L}{\partial\partial_\mu\phi}$$ and $$\mathcal L$$ is the Lagrangian density). We also know that this quantity can be thought as the generator of the transformation, meaning that, if we call $$F\equiv-\epsilon^\mu P_\mu$$, we have $$\delta_0\phi=\{\phi,F\} \Rightarrow \partial_\mu\phi=\{\phi,P_\mu\}$$

At this point I'm supposed to be able to say that $$\{P_\mu,P_\nu\}=0$$, but I don't know how to prove it with the means that I have. I think that, since $$P_\mu$$ does not depend on the particular point that I want to translate, it doesn't depend on $$x$$, so this could be a proof, but is it the only one? I'd like to prove it mathematically if it is possible.

Edit: In Roman, Introduction to QFT, page 77, the author says that also $$J_{\mu\nu}$$ (Lorentz generators) is coordinate-independent and he also uses this to prove the relation for $$\{J_{\mu\nu},J_{\sigma\rho}\}$$, and he claims that by saying that it follows from the definition (as he says for $$P_\mu$$), but

1. Once again, how this follows from the definition? Both $$P_\mu$$ and $$J_{\mu\nu}$$ are defined as an integral in $$d\sigma_\mu$$ over a space-like surface of a current density. But this is true for every generator, so are all generators always coordinate-independent? Do all of them commute with $$P_\mu$$? It doesn't seem true to me. For example, it is not true for $$\{P_\mu,J_{\sigma\rho}\}=g_{\mu\rho}P_\sigma-g_{\mu\sigma}P_\rho$$.
2. If, like Roman says, $$J_{\mu\nu}$$ is coordinate-independent, than $$\{J_{\sigma\rho},P_\mu\}=\partial_\mu J_{\sigma\rho}=0$$, but we know that $$\{J_{\sigma\rho},P_\mu\}=-\{P_\mu,J_{\sigma\rho}\}\neq 0$$, so how is that possible? Am I misunderstanding something?

$$\pi=\partial_0 \phi, \\ P^i = \int d^3x ~T^{0 i} (x)=\int d^3x ~\pi(x) \partial_i \phi(x) \\ P^0= \int d^3x ~(\pi^2 +(\partial_i \phi)^2 )/2,$$ so that, directly, given $$\{ \phi(x),\pi(y)\}=\delta^3(x-y)$$, $$\{ \phi(x), P^i\}= \partial_i \phi(x), \\ \{ \phi(x), P^0\}= \pi (x) =\partial_0 \phi(x), \\ \{ P^\mu, P^\nu \} = 0.$$