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Let $M$ be a Lorentzian spin $4$ manifold, i.e. admits a spin structure $Spin^+(M)\rightarrow M$, which is just a principal $Spin^+(1,3)$ bundle over $M$, which is compatible with the bundle of oriented, and time oriented, orthonormal frames over $M$, denoted $SO^+(M)$. Furthermore, let $P\rightarrow M$ be a principal $G$ bundle over $M$, where $G$ is a compact Lie group. We can then write the Yang-Mills-Dirac Lagrangian on $M$ as:

\begin{align*} \mathscr{L}_{YMD}[A,\Psi]=-\frac{1}{2}\langle F^A_M,F^A_M \rangle_{Ad(P)}-m^2\langle \Psi,\Psi\rangle_{S\otimes E}+\Re\left(\langle\Psi,D_A\Psi \rangle_{S\otimes E}\right). \end{align*} Here $F^A_M$ is the curvature two form of a gauge potential $A$, $Ad(P)$ is the adjoint bundle $Ad(P)=P\times _{Ad}\mathfrak{g}$, $S=Spin^+(M)\times _{\kappa}\mathbb{C}^4$ is the spinor bundle (where $\kappa$ is the spin representation), $E=P\times_\rho V$ is a vector bundle associated to $P$ (where $\rho$ is a representation of $G$ on $V$), making $S\otimes E$ a twisted spinor bundle, and $\Psi$ a gauge multiplet spinor field. Finally, $D_A$ is the twisted Dirac operator, which is given in a local gauge and orthonormal frame $e_i$ by \begin{align*} D_A\psi=\gamma^i\left(d\psi(e_i)+\frac{1}{4}\xi_{ab}(e_i)\gamma^{ab}\psi+\rho_*(A(e_i))\psi\right) \end{align*} where $\gamma^i$ is a gamma matrix, and $\gamma^{ab}=\frac{1}{2}[\gamma^a,\gamma^b]$. The one forms $\xi_{ab}$, are the one forms defining the Levi-Civita connection in this orthonormal frame, i.e. $$\nabla e_a=\xi_{ab}\eta^{ab}\otimes e_b.$$

I know that this is only a "classical" description of this picture, as I am not doing anything with path integrals, but I have noticed that two of the objects in $\mathscr{L}_{YMD}$ depend on the metric $g$, namely the Yang-Mills term, and the term which includes the twisted Dirac operator. Furthermore, when I pass to the action: \begin{align} S_{YMD}=\int_M \mathscr{L}_{YMD}[A,\Psi]\text{dvol}_g \end{align} where in coordinates $$\text{dvol}_g=\sqrt{-\det g}dx^1\wedge \cdots \wedge dx^4$$ each term will at least mildly depend on the metric if I vary the action with respect to $g$.

My question is then this: If I add the Einstein Hilbert action to the Yang-Mills-Dirac action, will this then describe a theory of gravity where the matter Lagrangian is $\mathscr{L}_{YMD}$? I know as a quantum theory, GR ends up being non-renormalizable, but does this set up describe the Lagrangian of that non renormalizable theory?

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  • $\begingroup$ Adding the EH action does not automatically yield a failed quantum gravity approach. Essentially, this will be a classical theory in a curved background if you don't quantize it, and a QFT in curved background if you use path integrals for the gauge field and the Dirac bispinors. If furthermore, you want to quantize gravity, good luck. $\endgroup$ Commented May 2, 2023 at 18:03
  • $\begingroup$ @JeanbaptisteRoux But the curved background will depend on the spinor fields and gauge field right? $\endgroup$
    – Chris
    Commented May 2, 2023 at 18:16
  • $\begingroup$ This is true, but quantizing a field does not quantize automatically another field that is in interaction with the first one. In QFT in curved space-time, we have the semiclassical EFE. So quantizing the gauge and fermion field but keeping the gravity side as classical will lead to an approximation to quantum gravity. For an introduction to QFT in curved space-time see Birrell & Davies $\endgroup$ Commented May 3, 2023 at 5:57
  • $\begingroup$ @JeanbaptisteRoux I have that book, just haven't gotten around to looking at it yet as it is at my home, and not where I go to school. I'll definitely take a look when I go back though. So does quantizing the gauge and fermion field actually lead to quantum effects influencing the curved background? Or is it essentially just treating the the quantum fields as classical, which then determine our curved background, and change the quantum effects because the metric is variable? $\endgroup$
    – Chris
    Commented May 3, 2023 at 6:51
  • $\begingroup$ Yes, this is the idea: by integrating out the gauge and fermion fields, you will have the generating functional of connected diagrams as an action, and you can find with it the semi-classical stress-energy tensor of the theory. You will have to renormalize it, though. $\endgroup$ Commented May 3, 2023 at 9:46

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