Does a good path integral exist in Loop Quantum Gravity?

The Hamiltonian operator of loop quantum gravity is a totally constrained system $$H = \int_\Sigma \mathrm{d}^3x\ (N\mathcal{H}+N^a V_a+G).$$

Here, $$\Sigma$$ is a 3-dimensional hypersurface; a slice of spacetime. Moreover, $$\mathcal{H}$$ is the Hamiltonian constraint, $$V_a$$ the diffeomorphism constraint, $$G$$ the Gauss law term and $$N,N^a$$ the corresponding constraint generators.

In research literature, this Hamiltonian was criticized to be not Hermitian and would not form a Lie algebra from its generators. The variables of the theory are Ashtekar's variable $$A_a^i$$ and the triad $$E_a^i$$. Therefore the Master constraint $$M:=\int_\Sigma \mathrm{d}^3x\ \frac{\mathcal{H}^2}{\sqrt{\det q}}$$ with 3-d-metric $$q_{ab}$$ was introduced that solves these issues. Loop quantum gravity can be treated canonically, but according to this paper, one can derive a path integral from the Master constraint. I can't understand the derivation of it (especially with the measure factor). Question: Is there a plausible path integral in 4-d-spacetime that computes spin foam amplitudes?

What if I treat Loop Quantum Gravity with the path integral with action $$S = \int \mathrm{d}^4x\ (E_a^i \dot{A_i^a}-N\mathcal{H}+N^a V_a+G) \tag{\star}$$ Is it plausible (this action is mentioned in one of my introductory textbooks) despite the non-hermiticity of the Hamiltonian? Or would this action lead to significant errors?

P.S. Is the path integral $$\int \mathrm{d}[E_a^i] \mathrm{d}[A_i^a] \mathrm{d}[N_{\text{Master}}] \exp\left(i E_a^i \dot{A_i^a} - i\int \mathrm{d}t N_{\text{Master}} M\right) \\ = \int \mathrm{d}[E_a^i] \mathrm{d}[A_i^a] \exp\left(i E_a^i \dot{A_i^a}\right) \delta(M)$$ a better version of the path integral induced by the action $$(\star)$$?

• May be related: there is a covariant approach to Loop Quantum Gravity called the spinfoam formalism. Spinfoams are based on path integrals. They is still under active research, and it is yet unclear whether they give the same results as canonical LQG or not. Commented Dec 14, 2016 at 6:02