7
$\begingroup$

Note: I don't know much about QFT, aside from some basics, I am a student and first and foremost a student of GR.

As far as I am aware, the fact that interactions are gauge theories allow some pretty natural ways of deriving them. For example, for QED, if one is given a matter field $\phi$ (matter here can be bosons as well), whose lagrangian is invariant under global $U(1)$ transformations, then demanding that this lagrangian should be invariant under $U(1)$ transformations that depend on spacetime points as well naturally leads to the introduction of an $U(1)$ gauge connection, and then conjuring up a gauge-invariant lagrangian for the gauge connection too gives one a system of Maxwell's equations being coupled to the dynamics of a matter field.

Since we know from classical electrodynamics that only charged particles interact electromagnetically, this also gives a natural way of telling which particle is charged. A particle is charged if the corresponding matter field admits a global $U(1)$ symmetry.

For example the complex scalar field $$\mathcal{L}_{KG\ (C)}=\partial_\mu\phi^\dagger\partial^\mu\phi-m\phi^\dagger\phi$$ is charged but the real version of the same thing isn't.


Now, assume that one does not know about Riemannian geometry or general relativity, but because all other interactions arise via a gauge principle like in QED, this person tries to give rise to a theory of gravity the same way.

One knows that gravity interacts with everything, so one cannot use some fancyful gauge group $G$, because every matter field needs to know this symmetry. One fundamental requirement of relativistic field theories is that the action should be Lorentz-invariant, so the only Lie group that satisfies this requirement is the Lorentz group, therefore one needs to gauge the Lorentz group.

The Lorentz group is however an "external" symmetry group, not an "internal" one, eg. Lorentz transforms are related to spacetime geometry.

It is clear that the usual way of doing stuff, namely to replace $\partial_\mu$ with some gauge connection $D_\mu$ won't work here. For example in SR, the coordinates $x^\mu$ form a 4-vector. What is the meaning of a position dependent transform $\Lambda^\mu_{\ \nu}(x)x^\nu$? Nothing.

Of course, if one does know Riemannian geometry and GR, one can see that you need to give up flat spacetime one way or another. The preferred way to do that (preferred in the sense of following the example of QED or QCD) is to make Lorentz symmetry "internal" and have every Lorentz tensor $S_{\mu\nu...}$ replaced with some "section" $S_{a,b...}$ and provide a function $\theta^a_{\mu}$ that relates the "internal" and "external" spaces to one another (a vielbein, basically, but I am using its interpretation as a solder form here) and also replace the volume element $d^4x$ with the invariant volume element $\det(\theta)d^4x$ and all "spacetime" indices $\mu,\nu$ should be interpreted as referring to a general coordinate system, and of course, one needs to introduce a gauge connection $D_\mu=\delta^a_b\partial_\mu+\omega_{\mu\ \ b}^{\ a}$ instead of $\partial_\mu$, but these steps do not follow naturally from the idea of making Lorentz transformations local.

Question: Assuming one does not know GR or differential geometry, but has the bright idea of creating a gauge theory of gravity by gauging the Lorentz group, is there any "natural" way of performing this gauging that leads to a consistent field theory of gravity that couples to everything?

If so, is this necessarily unique? I guess not, since it is not a fundamental requirement to have the gauge connection determined uniquely by the vielbein (Einstein-Cartan theory for example).

But is it necessary to have a vielbein actually appear? Is it possible to arrive at this purely from the idea of gauging the Lorentz group?

$\endgroup$
2
$\begingroup$

I know this probably doesn't address the whole question, but nevertheless I thought it might be of help to you.

As I understand, you would like to explore the possibility of unifying the frame $e$ with the spin connection $\omega$. There have certainly been attempts to do that.

Disclaimer: everything in this answer is speculative and unconfirmed by experiment. We are talking about theoretically possible approaches to unification.

Intro: Palatini/Holst gravity

First, consider the traditional frame-connection formulation of GR given by Palatini:

$$ S_{\text{Palatini}} [e, \omega] = \frac{1}{2 \kappa} \varepsilon_{IJKL} \int e^{I} \wedge e^{J} \wedge F^{KL}, $$

where $e^I = e^I_{\mu} dx^{\mu}$ is the frame 1-form and $\omega^{IJ} = \omega^{IJ}_{\mu} dx^{\mu}$ is the Lorentz algebra-valued spin connection and $F^{IJ} = d\omega^{IJ} + \omega^{IK} \wedge \omega_K^{\;J}$ is its curvature.

This action gives the compatibility condition $De=0$ when variated wrt $\omega$ ($D$ is the Lorentz-algebra covariant derivative) and Einstein's equations for $g_{\mu \nu} = e_{\mu}^I e_{\nu}^J \eta_{IJ}$ upon variation wrt $e$.

This can be modified by a topological Holst term as follows:

$$ S_{\text{Holst}} [e, \omega] = \frac{1}{2 \kappa} \int e^{I} \wedge e^{J} \wedge \left( \varepsilon_{IJKL} F^{KL} + \frac{1}{\gamma} F_{IJ} \right). $$

Classical equations of motion are independent of $\gamma$. However for $0\lt \gamma \lt \infty$ this theory can be consistently quantized with help of the spin network formalism to give Loop Quantum Gravity.

MacDowell-Mansouri gravity

MacDowell&Mansouri managed to pack the frame $e$ and the spin connection $\omega$ together as parts of the single $\mathfrak{so}(4,1)$-valued connection $A$. Or, for the negative cosmological constant (which isn't physically favored), the connection is $\mathfrak{so}(3,2)$-valued.

Have a look at this review.

Tombstone of gauge-gravity unification: the Coleman-Mandula theorem

This is a no-go theorem which states that any quantum field theory with a mass gap and with analytic S-matrix can't contain the Poincare group and the gauge group combined in any but trivial way.

Therefore, you can't unify gravity with gauge interactions in any nontrivial Lie group.

This theorem serves as one of the many justifications of the search for supersymmetry, since it doesn't hold in case of supersymmetric Lie algebras.

But there are pitfalls. First, QFTs might be only valid up to a certain energy scale (presumably, the scale of breaking of diffeomorphism symmetry, at which gravity separates from other interactions). Second, if we are in the de'Sitter space with positive cosmological constant, the theorem no longer holds.

Lisi's Clifford connection and $E8$

Garret Lisi has put out a different approach for unifying the frame and spin connection in this paper. The idea is to consider a single $Cl(3,1)$-valued connection and interpret $e$ and $\omega$ as grade-1 and grade-2 parts of this connection.

He also interprets physical fermions as BRST ghosts which arise after gauge-fixing.

Later, in his famous paper, he attempts to unify these with the gauge groups of the Standard Model as parts of the $E8$ gauge group.

Personally, I don't understand how the dynamics of the $E8$ theory works. It seems to me that the action for the model is not even $E8$-invariant.

$\endgroup$
1
$\begingroup$

You don't need QFT to gauge the Lorentz group and get GR (or its natural generalisation: Einstein-Cartan theory) from SR.

To define the Lorentz group, you need a local reference frame (i.e. the vieilbein, or tetrad), representend by four orthogonal 4-vectors $\boldsymbol{e}_a$ and their four dual covectors $\boldsymbol{e}^a$ ($\equiv \boldsymbol{e}^{* a}$), which must satisfy the following relations, as in special relativity ($\circ$ is the scalar product of 4-vectors and duals): \begin{align}\tag{1} \boldsymbol{e}_a \circ \boldsymbol{e}_b &= \eta_{ab}, &\boldsymbol{e}^a \circ \boldsymbol{e}^b &= \eta^{ab}, &\boldsymbol{e}_a \circ \boldsymbol{e}^b &= \delta_a^b. \end{align} Under any passive Lorentz transformation, the frame changes: \begin{equation}\tag{2} \tilde{\boldsymbol{e}}^a = {\Lambda_b}^a \, \boldsymbol{e}^b, \end{equation} where ${\Lambda_b}^a$ are constant matrix elements, as in special relativity. Gauging that group implies that these elements become position dependant. You then have to change the partial derivatives of "flat tensors" to some covariant derivative under local Lorentz transformation. This implies that you'll have to introduce a connection (the spin connection).

Take note that it is false to say that the coordinates $x^a$ define a 4-vector in special relativity. Even in SR, they are just coordinates (and you're even free to use curvilinear coordinates $x^{\mu}$ in SR, like spherical coordinates, etc, and not just cartesian coordinates $x^a$). It is the orthogonal tetrad which should be used, or the infinitesimal cartesian coordinates change: \begin{equation}\tag{3} d\tilde{x}^a = {\Lambda^a}_b(x) \, dx^b. \end{equation} Here, the "infinitesimal" $dx^a$ is actually a representation of the dual base: $dx^a \equiv \boldsymbol{e}^a$.

The rest (connection, curvature, ...) becomes pretty complicated very fast!

For more details on all this, you should read Weinberg's book on GR.

$\endgroup$
0
$\begingroup$

If you want to write down a gravity action a la Yang-Mills (note that there is no Hodge dual $*$): $$ S \sim \int F\wedge F, $$ you have to extend the local Lorentz group to (anti) De Sitter group (cf. MacDowell-Mansouri), which combines vielbein/tetrad one-form $e$ and Lorentz connection (a.k.a. spin connection) one form $\omega$ into a unified connection one form of (anti) De Sitter gauge group $$ A = e + \omega. $$ And the curvature two-form $F$ corresponding to the (anti) De Sitter gauge group follows the same Yang-Mills definition (for abbreviation, Lorentz indices are not shown here and exterior products between differential forms are assumed): $$ F = dA + A^2 = (d\omega + \omega^2) + (de + e\omega + \omega e) + e^2 $$ where $(d\omega + \omega^2)$ is the spin connection curvature (Riemann) tensor, $(de + e\omega + \omega e)$ is torsion tensor.

Nevertheless, the MacDowell-Mansouri's version of gravity has several drawbacks:

  1. First of all, the (anti) De Sitter group has to be broken to arrive at the remaining unbroken Lorentz local symmetry. The (anti) De Sitter symmetry breaking mechanism is put in by hand in MacDowell-Mansouri's original paper. Other following-up papers of spontaneous symmetry breaking mechanism are rather ad hoc.
  2. The (anti) De Sitter connection (specifically the vielbein/tetrad portion) couples left-handed fermions to right-handed fermions. It spells trouble for the standard model chiral fermions.
  3. The gauge Lagrangian of MacDowell-Mansouri's version of gravity would involve a cosmological constant $e^4$ which is of the order of the Planck scale, which is also troublesome.
  4. To afford a supersymmetric version of the MacDowell-Mansouri's gravity, one has to leverage an accidental isomorphism between anti-de Sitter $\mathfrak{so}(2,3)$ and $\mathfrak{sp}(4,\mathbb{R})$. Hence the super-gravity/superstring saga is confined to AdS with a negative cosmological constant and confronted by the "swampland" predicament.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.