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Suppose I have a 2-form field $B$ and a Lagrange multiplier field $\lambda$, then the Lagrangian

$S = \int (B \wedge \delta B + \lambda \delta B \wedge \delta B)$

with a Lie derivative operator $\delta$ (doesn't change the differential form) generates a probability for scattering with the fields $B$ (see also: http://thewinnower.com/papers/2658-topological-dipole-field-theory).

Now suppose I have (EXAMPLE ONLY!) to compute the expectation value (integral measure does also contain the normalization factors):

$<B_{\alpha_1 \beta_1}(x_1)B_{\alpha_2 \beta_2}(x_2)B_{\alpha_3 \beta_3}(x_3)B_{\alpha_4 \beta_4}(x_4)>= \int d[B] \times \int d[\lambda] B_{\alpha_1 \beta_1}(x_1)B_{\alpha_2 \beta_2}(x_2)B_{\alpha_3 \beta_3}(x_3)B_{\alpha_4 \beta_4}(x_4)e^{iS}$.

In k-space I can write for the action $S = (2 \pi)^4\int d^4k \epsilon_{\alpha \beta \mu \nu}B_k^{\alpha \beta}\delta_k B_k^{\mu \nu} + (2 \pi)^4 \int \int d^4k d^4k' \lambda_{k'-k}\epsilon_{\alpha \beta \mu \nu}\delta_{k'}B_{k'}^{\alpha \beta}\delta_k B_k^{\mu \nu}$.

My question is the following: Since $\lambda(x)$ is a Lagrange multiplier field one has a scattering term of the action given by $S_{sc} = (2 \pi)^4 \int \int d^4k d^4k' \lambda_{k'-k}\epsilon_{\alpha \beta \mu \nu}\delta_{k'}B_{k'}^{\alpha \beta}\delta_k B_k^{\mu \nu}$. The other term is the kinetic term (Generalized gaussian integral).

Expanding the scattering term in Taylor series one obtains the elementary scattering processes: $B_{k'}^{\alpha \beta} \mapsto B_{k'}^{\mu \nu}$. There was transferred an excess of energy-momentum $\lambda_{k-k'}$. Has the constraint generated a nonconservation of energy and momentum?

If one knows it (this must not be answered necessarily): Loop quantum gravity is a QFT with constraints. Is in Loop quantum gravity energy and momentum conserved?

Every reply would be greatly appreciated.

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    $\begingroup$ Please substantiate the claim "In k-space I can write for the action", especially, explain what $\delta_{k'}$ it. Be mindful that the $\delta$ from the article you cite is a Cech boundary operator, not an actual derivative, and the $B$ a representant of a cohomology class. $\endgroup$
    – ACuriousMind
    Sep 26, 2015 at 22:27
  • $\begingroup$ Yes, it is a Cech coboundary operator in the article. $\endgroup$
    – kryomaxim
    Sep 26, 2015 at 22:30
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    $\begingroup$ Dear kryomaxim: Are you in any way associated with the author of the linked article? For your information, Physics.SE has a policy that it is OK to cite oneself, but it should be stated clearly and explicitly in the post itself, not in attached links. $\endgroup$
    – Qmechanic
    Sep 26, 2015 at 23:35
  • $\begingroup$ sorry; I haven't seen this rule. $\endgroup$
    – kryomaxim
    Sep 26, 2015 at 23:50

1 Answer 1

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Suppose that the field $B $ can be decomposed into $B=gB'+\phi \delta B$ with the coupling constant $g $ and $\delta \phi = I$ where $ \phi$ is indicator function on maximum set intersection such that $\delta \phi $ lies completely on the support of $\delta B $.

Now the integration haar measure can be decomposed into $d [B]=d [B'] d [\delta B \phi]$. After substitution of all that into the expectation value one can perform integration over $B'$ and get delta distributions. Therefore, integration over $\delta B \phi $ can be performed.

Finally, $\lambda $ integrations can be done and one gets a number (the lambda integration should be regularized by making integration cutoff).

This number is independent on energy and momentum; hence energy and momentum is conserved in the theory.

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