Replacing $U(1)$ covariant derivative with $GL(4,\mathbb{R})$ covariant derivative... does it give quantum gravity?

Question

I realize that many questions about deriving quantum gravity have been asked multiple times before on this forum, but it hasn't been asked exactly like I am doing here. I would like to know what specifically I get with this derivation; quantum gravity, quantum mechanics on curved space, something else, nothing? Also, if there are problems with it, what are they exactly --- non-renormalizable, transformations of GR violate the equations?

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If I define an action as

$$ \mathcal{S}=\int \bar{\psi} (i\hbar c \gamma^\mu D_\mu - m c^2)\psi-\frac{1}{4 \mu_0} F_{\mu\nu}F^{\mu\nu} $$

Then, this is QED.

What if I gauge the wavefunction with respect to a general linear transformation:

$$ \psi'=g\psi g^{-1} $$

Then, I get the following gauge

$$ D_\mu \psi = \partial_\mu \psi -[iqA_\mu, \psi] $$

but, since the gauge is general linear, the field is:

$$ R_{\mu\nu}=[D_\mu,D_\nu] $$

Consequently, if I write the following action:

$$ \mathcal{S}=\int \bar{\psi} (i\hbar c \gamma^\mu D_\mu - m c^2)\psi-\frac{1}{4} R_{\mu\nu}R^{\mu\nu} $$

Is it quantum gravity. What are the problems with it?

Answer 1

There are only two small problems: this approach has nothing to do with gravity, and it is not at all quantum :)

Firt, your action doesn't describe gravity; it describes Yang-Mills theory with the group $GL(n) = U(1) \times SL(n)$. Not gravity.

There's a formulation of gravity in the gauge theory language but it uses a different action:

$$ S[e, A] = \int d^4 x | \det e | e_a^{\mu} a_b^{\nu} F^{ab}_{\mu \nu} $$ with $A$ a $SO(3,1)$ connection, $F$ its curvature tensor, and $e$ the tetrad field that maps the tangent space to a point in spacetime to the abstract space $R^4$ and is invertible by definition.

You can pass to ordinary variables by $$ g_{\mu \nu} (x) = \eta_{a b} e^a_{\mu} (x) e^b_{\nu} (x). $$

To couple to fermions, replace $\partial_{\mu}$ by the covariant derivative that acts on objects in the spinor representation of $SO(3,1)$.

The second problem is that nothing about this is quantum. This is a completely classical theory, as classical as they come. What makes you think this is a quantum theory?