# How to derive infinitesimal gauge transformations from constraints?

I am reading some papers about quantizing the gravitational fields, for example, here, here, and here. Since the classical actions for gravitational fields are singular, they contain some constraints. I have no problem in finding these constraints.

In the ADM form, the primary constraints are $$\pi_0=0$$ and $$\pi_i=0$$. $$\pi_0$$ and $$\pi_i$$ are the canonical momenta conjugate to the lapse function $$N$$, and the shift vector $$N^i$$. Also, the spatial metric is approximated by $$g_{ij}=\delta _{ij}+2\kappa \gamma_{ij}$$. From the primary constraints we can have the secondary constraints, which are $$C_0=\partial^i \partial^j \gamma_{ij}-\partial^2 \gamma=0,$$ $$C_i=-2 \partial^j p_{ij}=0$$ where $$p_{ij}$$ are the momenta conjugate to the metric $$h_{ij}$$.

The papers then read that the combination $$C(\lambda_0,\lambda_i)=\int d^3 x [\lambda_0 (x) C_0(x)+\lambda^i(x)C_i(x)]$$ generates the infinitesimal gauge transformations $$\delta \gamma_{ij}=\frac 1{2\kappa} (\partial_i \lambda_j +\partial_j \lambda_i) ,$$ $$\delta p^{ij}=\frac 1 {2\kappa} (\delta^{ij} \partial^2 \lambda_0-\partial^i \partial^j \lambda_0) .$$

I have read some textbooks about constrained Hamiltonian. But I still could not understand how these infinitesimal gauge transformations are derived.

Looking for some hints.

• Do you know the Dirac conjecture, which states that gauge transformations are generated by first class constraints? Dec 20, 2021 at 12:45
• Are you confused about how the specific combination arises, or do you not understand what it means for it to "generate" the transformations? (In case it is the latter: "generate" in the Hamiltonian context always means "by taking the Poisson bracket with it") Dec 20, 2021 at 12:53
• Thanks! Qmechanic and ACuriousMind. The Dirac conjecture does lead to the procedure mentioned by ACuriousMind. I will try to answer this question myself. Dec 21, 2021 at 1:12

According to Dirac's conjecture, the first-class constraints $$\Omega_A$$ will lead to the following symmetry transformations, (neglecting the time parameters) $$\delta q^i=\epsilon^A \{q^i,\Omega_A\},~~ \delta p_i=\epsilon^A \{p_i,\Omega_A\}.$$
Also, since divergence terms vanish, the combination of constraints can be written as $$C(\lambda_0,\lambda_i)=\int d^3 x [\lambda_0 (x) C_0(x)+\lambda^i(x)C_i(x)] =\int d^3 x[\gamma_{ij}\partial^i \partial^j \lambda_0 -\delta^{ij}\gamma_{ij}\partial^2 \lambda_0 +2 p^{ij}\partial_j \lambda_i] .$$
(I am not very sure about the following step.) Then according to Dirac's conjecture, $$\delta g_{ij}=\{g_{ij}, \gamma_{ij}\partial^i \partial^j \lambda_0 -\delta^{ij}\gamma_{ij}\partial^2 \lambda_0 +2 p^{ij}\partial_j \lambda_i\} =2(\partial_j \lambda_i+\partial_i\partial_j)$$ and hence $$\delta \gamma_{ij}=\frac 1 \kappa (\partial_j \lambda_i+\partial_i\partial_j).$$ It differs from the above equation by a factor of $$1/2$$, which is not significant and can be absorbed into $$\lambda_i$$.
The same goes for $$\delta p^{ij}$$.