# Canonical Commutation relations in gravity

The canonical commutation relations in gravity are sometimes written $$[\gamma_{ij}(x),\pi^{kl}(y)]=\frac{i\hbar}{2}(\delta_i^k\delta_j^l+\delta_i^l\delta_j^k)\delta^3(x-y),\tag{0}$$ where $$\gamma_{ij}$$ is the 3-metric. This means \begin{align} [\gamma_{11}(x),\pi^{11}(y)]&=i\hbar\delta^3(x-y) \tag{1} \\ [\gamma_{12}(x),\pi^{12}(y)]&=\frac{i\hbar}{2}\delta^3(x-y) \tag{2} \end{align} for example.

Eq. (1) looks standard so I would conclude that $$\gamma_{11}$$ is canonically conjugate (c.c.) to $$\pi^{11}$$.

Eq. (2), however, has the factor of $$\frac{1}{2}$$ which doesn't look standard at all. It would seem to suggest that $$\gamma_{12}$$ and $$\pi^{12}$$ are not canonically conjugate. Any comments?

EDIT: I think the simple conclusion is that $$\gamma_{ij}$$ is c.c. to $$2\pi^{ij}$$ when $$i\neq j$$ and c.c. to $$\pi^{ij}$$ when $$i=j$$.

• Can you ask a more specific question than "any comments"? This is far from my specialty but I see nothing particularly worrisome about factors of 1/2 appearing in the context of symmetric tensors ($\gamma_{ij} = \gamma_{ji}$, right?) Jul 14 at 18:00

1. The factor $$\frac{1}{2}$$ in the symmetrized Poisson bracket relations \begin{align} \{\gamma_{ij}(x),\pi^{kl}(y)\} ~=~&\frac{1}{2}(\delta_i^k\delta_j^l+\delta_i^l\delta_j^k)\delta^3(x-y),\cr \{\gamma_{ij}(x),\gamma_{kl}(y)\}~=~&0,\cr \{\pi^{ij}(x),\pi^{kl}(y)\}~=~&0, \end{align} \tag{A} is a natural consequence of the symmetry of the metric tensor $$\gamma_{ij}$$ (and the symmetry of $$\pi^{kl}$$), cf. e.g. this related Phys.SE post.

2. We may write the Schrödinger representation as $$\pi^{kl}(x)~=~\frac{\hbar}{2i}\left(\frac{\delta}{\delta\gamma_{kl}(x)}+\frac{\delta}{\delta\gamma_{lk}(x)} \right).\tag{B}$$

3. Alternatively, we may view the tensors $$\gamma_{ij}$$ and $$\pi^{kl}$$ as fundamentally non-symmetric tensors, and proceed as follows:

• Option (i): They satisfy the symmetrized Poisson bracket relation (A). Only later impose the first-class constraints \begin{align} \phi_{ij}~:=~&\gamma_{ij}-\gamma_{ji}~\approx~0,\cr \chi^{kl}~:=~&\pi^{kl}-\pi^{lk}~\approx~0,\end{align}\tag{C} where $$\{\phi_{ij}(x),\chi^{kl}(y)\} ~\stackrel{(A)+(C)}{=}~0.\tag{D}$$

• Option (ii): They satisfy the non-symmetrized Poisson bracket relations \begin{align} \{\gamma_{ij}(x),\pi^{kl}(y)\} ~=~&\delta_i^k\delta_j^l\delta^3(x-y),\cr \{\gamma_{ij}(x),\gamma_{kl}(y)\}~=~&0,\cr \{\pi^{ij}(x),\pi^{kl}(y)\}~=~&0. \end{align} \tag{E} Then $$\{\phi_{ij}(x),\chi^{kl}(y)\} ~\stackrel{(C)+(E)}{=}~2(\delta_i^k\delta_j^l-\delta_i^l\delta_j^k)\delta^3(x-y).\tag{F}$$ Only later impose the second-class constraints (C). The Dirac bracket becomes \begin{align}\{F,G\}_D ~=~&\{F,G\}\cr ~-~&\frac{1}{4}\int\!d^3z\{F,\chi^{mn}(z)\}\{\phi_{mn}(z),G\}\cr ~+~&\frac{1}{4}\int\!d^3z\{F,\phi_{mn}(z)\}\{\chi^{mn}(z),G\}. \end{align}\tag{G}
One may check that the Dirac bracket (G) reproduces the symmetrized Poisson bracket relations \begin{align} \{\gamma_{ij}(x),\pi^{kl}(y)\}_D ~=~&\frac{1}{2}(\delta_i^k\delta_j^l+\delta_i^l\delta_j^k)\delta^3(x-y),\cr \{\gamma_{ij}(x),\gamma_{kl}(y)\}_D~=~&0,\cr \{\pi^{ij}(x),\pi^{kl}(y)\}_D~=~&0. \end{align} \tag{H}

Both options (i) & (ii): The variables $$\gamma_{ij}$$ and $$\pi^{kl}$$ are then canonically conjugate variables on the constrained submanifold, cf. OP's question.

• Yes but then I can't conclude that $\gamma_{12}$ is canonically conjugate (c.c.) to $\pi^{12}$ (because for $q$ and $p$, say, to be c.c. they must satisfy $[q,p]=i$). Jul 14 at 18:04
• I updated the answer. Jul 14 at 19:45