# What is the commutator of a Poisson bracket and the covariant derivative?

Consider a classical vector field $V^\mu$ on a curved background. We make a 3+1 split of coordinates into $t,x^i$, where $x^i$ are coordinates on spatial hypersurfaces $\Sigma$ and $t$ the parameter labeling them.

Now consider a canonically conjugate $$\tilde{\pi}_\mu = \frac{\partial \tilde{\mathcal{L}}}{\partial (\partial_0 \phi^\mu)},$$ where $\tilde{\mathcal{L}}$ is the Lagrangian scalar (the Lagrangian density would be $\mathcal{L} = \tilde{\mathcal{L}} \sqrt{-g}$). Then the following Poisson bracket holds $$\{V^\alpha(x^i,t),\tilde{\pi}_\beta (y^i,t)\} = \frac{1}{\sqrt{d}}\delta^{\alpha}_{\;\beta} \delta^{(3)} (x^i - y^i)$$ where $\sqrt{d}$ is the square root of the determinant of the metric $d_{ij}$ induced on $\Sigma$.

Now consider the total momentum $$P_\mu(t) \equiv \int_\Sigma (\tilde{\pi}_\alpha(x^i,t) V^{\alpha}_{\;\; ;\mu}(x^i,t) - \delta^t_\mu \tilde{\mathcal{L}}(x^i,t)) \sqrt{d} \,d^3\! x$$ where $V^\alpha_{\;\;;\mu} = \nabla_\mu V^\alpha$ is the covariant gradient and $d\Sigma = \sqrt{d} \, d^3 \! x$ is a covariant spatial volume element on $\Sigma$.

I would now like to evaluate brackets such as $\{V^\alpha(y^i,t), P_\mu (t)\},\{P_\mu,P_\nu\}$ or $\{V^{\alpha}_{\;\; ;\mu},P_\nu\}$ to explore this algebra further. The problem is, however, that the Poisson bracket and $\nabla_\mu$ obviously do not commute because if I swap their order in different ways, I seem to get different results in every case. So, how does one get the covariant derivative outside the Poisson bracket?

In other words, I am looking for $A^\alpha_{\; \beta \mu}$ and $B^\alpha_{\; \beta \nu}$ such that $$\{V^\alpha_{\;\;;\mu}(x^i,t), \tilde \pi_\beta(y^i,t)\} = \nabla^{(x)}_\mu\{V^\alpha (x^i,t), \tilde \pi_\beta(y^i, t)\} + A^\alpha_{\; \beta \mu}$$ $$\{V^\alpha(x^i,t), \tilde \pi_{\beta;\nu}(y^i,t)\} = \nabla^{(y)}_\nu\{V^\alpha (x^i,t), \tilde \pi_\beta(y^i, t)\} + B^\alpha_{\; \beta \nu}$$ What are they and how can I find them?

• You differentiate with respect to $x$ and $y$, so there should no problem to write the derivative in front of the anticommutator (since the other argument depends on another space-time point). You only have to look at the case where $x=y$. Jul 28, 2017 at 9:19
• @Alpha001 Well, I am working through it, one thing I neglected previously is that $\sqrt{d}$ is not constant, and that $V^{\mu;0}$ should actually be written as a function of $\pi^\mu$. I will maybe post this as a self-answered question once the dust settles after the weekend but if you want to please post an answer now.
– Void
Jul 28, 2017 at 10:17
• Out of curiosity, what is/are the Hamiltonian and/or Lagrangian of the theory? Jul 29, 2017 at 14:51
• @Qmechanic The point is to derive the relations independent of the Lagrangian. I assume that you were asking because of the curious commutation relations etc., I have corrected these issues in the answer.
– Void
Aug 28, 2017 at 16:38

Much of the difficulties with the computation come from the fact that the "proper" canonically conjugate momentum is actually defined as $$\pi_\mu = \frac{\partial (\tilde{\mathcal{L}} \sqrt{-g})}{\partial (\partial_0 V^\mu)} = \frac{\partial \mathcal{L}}{\partial (\partial_0 V^\mu)}\,.$$ This momentum $\pi_\mu = \tilde{\pi}_\mu \sqrt{-g}$ is then itself a vector density on $\Sigma$ and the Poisson bracket reads $$\{V^\alpha(x^i,t), \pi_\beta (y^i,t)\} = \delta^\alpha_\beta \delta^{(3)}(x^{i} - y^i)\,.$$ Note that in the original question the $\{V^\alpha(x^i,t), \tilde \pi_\beta (y^i,t)\}$ bracket is stated incorrectly, because we see that we have $$\{V^\alpha(x^i,t), \tilde \pi_\beta (y^i,t)\} = \frac{1}{\sqrt{-g}}\delta^\alpha_\beta \delta^{(3)}(x^{i} - y^i)\,.$$ Using the properties of the delta function it is then easy to show that the following Poisson bracket can be defined for the functionals of the fields: $$A(t)[\pi,\phi] = \int \mathcal{A}(\pi_\alpha,\pi_{\alpha,i}, V^\alpha, V^\alpha_{\;,i},x^i,t) d^3 x$$ $$B(t)[\pi,\phi] = \int \mathcal{B}(\pi_\alpha,\pi_{\alpha,i}, V^\alpha, V^\alpha_{\;,i},x^i,t) d^3 x$$ $$\{A(t),B(t)\} = \int \frac{\delta \mathcal{A}}{\delta V^\alpha} \frac{\delta \mathcal{B}}{\delta \pi_\alpha} - \frac{\delta \mathcal{B}}{\delta V^\alpha} \frac{\delta \mathcal{A}}{\delta \pi_\alpha} d^3 x$$ where $\mathcal{A},\mathcal{B}$ are densities on $\Sigma$ and $\delta \mathcal{F}/\delta f$ is the variational derivative \begin{equation} \frac{\delta \mathcal{F}}{\delta f} = \frac{\partial \mathcal{F}}{\partial f} - \frac{\partial \;}{\partial x^i} \frac{\partial \mathcal{F}}{\partial (f_{,i})}\,, \end{equation} where we have assumed that $\mathcal{F}$ is dependent only on $f$ and its first-order gradients (for higher order gradients we get a series of analogous terms of varying sign). This addresses only spatial gradients on $\Sigma$ and not the temporal ones.
The temporal gradients can be eliminated by $field_{,0} = \{field,\mathcal{H}\}$, some first order gradients will be possible to eliminate by substituting $$\pi_\mu V^{\mu}_{\;,0} - \mathcal{L} = \mathcal{H}\,.$$
As for the commutators of gradients of fields, with these better variables it is easy to compute $$\{V^\alpha_{\;;\mu}(x^i,t), \pi_\beta (y^i,t)\} = (\delta^\alpha_\beta \partial_{\mu(x)} + \Gamma^\alpha_{\mu\beta})\delta^{(3)}(x^{i} - y^i)$$ $$\{V^\alpha(x^i,t), \pi_{\beta|\mu} (y^i,t)\} = (\delta^\alpha_\beta \partial_{\mu(y)} - \Gamma^\alpha_{\mu\beta})\delta^{(3)}(x^{i} - y^i)\,,$$ where the symbol $\pi_{\beta|\mu}$ stands for a pseudo-covariant derivative $\pi_{\beta|\mu} = \pi_{\beta,\mu} - \Gamma^\gamma_{\mu\beta} \pi_\gamma$ (remember that $\pi_\mu$ is not a covariant quantity) and of course $\partial_0 \delta^{(3)} = 0$. This derivative proves to be very useful in the computation of many brackets.