Poisson Bracket in General Relativity and tensor weight

Question

I'm a bit confused about the tensor density weight of Poisson brackets in general relativity and their covariance. It's perhaps related to being unclear as to what happens when I integrate a scalar density of some weight other than 1. Let's say I have the Poisson bracket of General Relativity in the 3+1 ADM formalism acting on some local scalar $f(x)$ on a space-time slice and some scalar quantity $G$. ($G$ could be the Hamiltonian, and $f(x)$ could be a scalar, but it could also be a scalar density, e.g .$\sqrt{g}$ which changes things but not the essence of what I'm asking). The Poisson bracket is given by \begin{align} \{f(x),G\}&=\int d^3y \Big[ \frac{\delta f(x)} {\delta g_{ab}(y)}\frac{\delta G}{\delta \pi^{ab}(y)} - \frac{\delta f(x)} {\delta \pi^{ab}(y)}\frac{\delta G}{\delta g_{ab}(y)}\Big] \\ &\stackrel{?}{=} \frac{\delta f(x)} {\delta g_{ab}(x)}\frac{\delta G}{\delta \pi^{ab}(x)} - \frac{\delta f(x)} {\delta \pi^{ab}(x)}\frac{\delta G}{\delta g_{ab}(x)} \end{align} with $g_{ab}$ the 3-metric and using the convention of taking its conjugate momenta $\pi^{ab}$ a tensor density of weight one (since we derive it from the Lagrangian density). 2 questions: The first is that the tensor weight of the first expression seems to be -2 (plus whatever comes with $f$, since I have the $d^3y$ on the top and the $\delta \pi^{ab}$ on the bottom. Since the left hand side is usually something like $\partial_tf(x)$, I would have expected it to have a tensor weight of 1. And this expression doesn't look like it will give diffeomorphism invariance, although I accept that it must (I guess one needs to consider how the 3-manifold sits in the 4-manifold for this).

There is some discussion of the invariance properties of the Poisson bracket here: Poisson brackets in curved spacetime, but I don't find it particularly enlightening. Anyone have a simple explanation?

Answer 1

1. This is not limited to GR. More generally, given an $(r,s)$ tensor field $\phi(x)$, the conjugate momentum field $\pi(x)$ is an $(s,r)$ tensor density field. See also this related Phys.SE post$^1$.

2. Given two scalar local functionals of the form $$ F =~ \int d^3x~\rho(x)~f(x)\qquad\text{and}\qquad G =~ \int d^3x~\rho(x)~g(x),\tag{A}$$ where $\rho(x)$ is a density field, and $f(x),g(x)$ are scalar fields, then the functional derivatives$^2$ $$\frac{\delta F}{\delta\phi(x)}~=~\frac{\partial [\rho(x) f(x)]}{\partial \phi(x)} -\frac{d}{dx^i} \frac{\partial [\rho(x) f(x)]}{\partial [\partial_i\phi(x)]}+\ldots\tag{B}$$ and $$ \frac{\delta G}{\delta\pi(x)}~=~\frac{\partial [\rho(x) g(x)]}{\partial \pi(x)} -\frac{d}{dx^i} \frac{\partial [\rho(x) g(x)]}{\partial [\partial_i\pi(x)]}+\ldots \tag{C}$$ are an $(s,r)$ tensor density field and an $(r,s)$ tensor field, respectively. Therefore the canonical Poisson bracket $$\{ F,G\}~=~ \int d^3x~\left(\frac{\delta F}{\delta\phi(x)}\frac{\delta G}{\delta\pi(x)}-\frac{\delta F}{\delta\pi(x)}\frac{\delta G}{\delta\phi(x)}\right)\tag{D}$$ is again a scalar local functional.

3. In particular,$^3$ $$\begin{align} \{ f(x),G\}&=~ \int d^3y~\left(\frac{\delta f(x)}{\delta\phi(y)}\frac{\delta G}{\delta\pi(y)}-\frac{\delta f(x)}{\delta\pi(x)}\frac{\delta G}{\delta\phi(y)}\right)\cr &=~ \frac{\delta f(x)}{\delta\phi(x)}\frac{\delta G}{\delta\pi(x)}-\frac{\delta f(x)}{\delta\pi(x)}\frac{\delta G}{\delta\phi(x)}\end{align} \tag{E}$$ in terms of 'same-space' functional derivatives $$ \frac{\delta f(x)}{\delta\phi(x)}~:=~ \frac{\partial f(x)}{\partial \phi(x)} -\frac{d}{dx^i} \frac{\partial f(x)}{\partial [\partial_i\phi(x)]}+\ldots\tag{F} ,$$ cf. e.g. my Phys.SE answer here.

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$^1$ If $\phi(x)$ is a tensor density field, then the conjugate momentum field $\pi(x)$ is a tensor field, i.e. then the roles are reversed.

$^2$ The ellipsis $\ldots$ denotes possible terms from higher-order space-derivatives.

$^3$ In this answer we use the convention that the Dirac delta distribution $\delta^3 (x,y)$ is density-valued $$\int d^3y~\delta^3 (x,y) f(y)~=~f(x). \tag{G}$$ Moreover, we use the convention that $$ \frac{\delta\phi(x)}{\delta\phi(y)}~=~\delta^3(x,y) \tag{H} .$$