At this point in a talk on supergravity, Lance Dixon says "Gravity = (gauge theory)$^2$". (I found a slightly different set of his slides online; the closest is slide 19 here.) He explains that, due to the graviton spin being twice that of a gluon, and because graviton scattering amplitudes are products of pairs of gluon amplitudes (which assists SUGRA calculations), a graviton can be thought of as a combination of two "gluons".

So how does this idea work? My suspicion is that the gluons in this analogy are the veilbein $e_\mu^A$, a set of $D$ vectors in a $D$-dimensional spacetime viz. $g_{\mu\nu}=e_\mu^A\eta_{AB}e_\nu^B$. Comparing this with the eight $A_\mu^a$ in QCD, I'm guessing we "square" a veilbein gauge group $G\le U\left( D\right)$ with $D$ generators. However, since we'd want $D^2$ pairs $e_\mu^Ae_\nu^B$, we cannot simply use $G^2:=G\times G$ as a Lie group (which would only have dimension $2D$). In any case, using $G^2$ as a Lie group would just give a gauge theory, not a "gauge theory$^2$" as Dixon suggests. In other words, we have to "square" the theory in a different sense.


2 Answers 2


I think I see the source of your confusion: when amplitudes people talk about gravity being gauge theory squared, we've already stripped off the gauge group.

You talk about eight $A_\mu^a$ in QCD. These can be written as $A_\mu^a=T^a e_\mu$, where $T^a$ is an element of the Lie Algebra $SU(3)$ and $e_\mu$ is the polarization vector. On-shell, this has $D-2$ degrees of freedom. If we want to get gravity, we strip off the color pieces and square, getting $e_\mu e_\nu=e_{\mu \nu}$, the polarization tensor for an on-shell graviton. So while your instinct in thinking about the veilbein was correct, in this case the construction is a bit simpler.

(Note that we are exclusively talking about on-shell states here. If you want to tell this story off-shell, things get a bit more complicated. Since this whole story come from scattering amplitudes, the external states are always on-shell, and you can often deal with on-shell internal states in intermediate stages of the calculation via generalized unitarity and the like, see chapter 6 here.)

Let's see that this gives the right number of degrees of freedom. For example, in four dimensions each gluon polarization can be plus or minus. $e_\mu^+ e_\nu^+=e_{\mu \nu}^+$ gives the positive helicity polarization of the graviton, while $e_\mu^- e_\nu^-=e_{\mu \nu}^-$ gives the negative helicity polarization.

What about $e_\mu^- e_\nu^+$ and $e_\mu^+ e_\nu^-$? Here it would seem that we have too many states. This turns out to be a feature, not a bug, because the theory we're generating here isn't pure gravity, it's axion-dilaton gravity. In four dimensions, the antisymmetric field $B_{\mu\nu}$ is dual to a pseudoscalar, the axion. Along with the metric trace, the dilaton, we have the right number of fields to match gauge theory squared on-shell. We get axion-dilaton gravity rather than pure gravity in part because these constructions were originally found in string theory and in supergravity theories, where in general you have axion-dilaton gravity. If I remember correctly, there are ways to modify the setup to kill the extra freedom and get back to pure gravity, but they're somewhat messy.

  • $\begingroup$ Do you know what does the "squaring" looks like at the Lagrangian level? For instance do I multiply two Dirac Lagrangian together : $(\overline{\psi} D_\mu \psi)(\overline{\psi} D_\mu \psi)$? $\endgroup$
    – Anon21
    Jul 8, 2022 at 13:29
  • $\begingroup$ Doing it at the Lagrangian level would mean doing it off-shell, and it's still not known whether that's possible. It's not literally squaring the Lagrangian terms, for the same reason the on-shell version isn't literally squaring the amplitudes: you don't square the propagators. If you want to see an example where this kind of thing works, here's a recent one you can look at: arxiv.org/abs/2112.11452 $\endgroup$
    – Matt
    Jul 9, 2022 at 16:57

The graviton is in a sense an entangled triplet state of two gauge bosons. The glueball state of two gluons in a triplet state is quantum mechanically the same as a graviton. However, it is not the case a glueball is a graviton. These decay quickly because the color charge internal to the system is very strong. The S-dual of QCD may though have a very weak color charge, which will make the ensuing entangled pair very weakly interacting and stable.

The two considerations, one a gauge transformation defined by the root vectors of a Lie algebra and the other gravitation, are considered with respect to each other. The Riemann curvature with vanishing Ricci curvature is the Weyl tensor. For sourceless region the curvature is purely vacuum and given by the Weyl tensor. Given null vectors $U^\alpha$ the null vector $U^{\alpha\beta}~=~[U^\alpha,~U^\beta]$ is defined. These are eigen-bivectors of the Weyl tensor $$ \frac{1}{2}{C^{\mu\nu}}_{\alpha\beta}U^{\alpha\beta}~=~\lambda U^{\mu\nu}. $$ \vskip.1in Consider the metric composed of the null vectors $x_\alpha,~y_\alpha,~z_\alpha$, wuch that $z^\alpha y_\alpha~=$ $x^\alpha y_\alpha~=~-1$ and $\bar y^\alpha y_\alpha~=~1$, $$ g_{\alpha\beta}~=~x_\alpha z_\beta~+~z_\alpha x_\beta~+~y_\alpha\bar y_\beta~+~\bar y_\alpha y_\beta. $$ There are then three possible null bivectors $$ U_{\alpha\beta}~=~-X_{[\alpha}y_{\beta]},~V_{\alpha\beta}~=~z_{[\alpha}y_{\beta]},~W_{\alpha\beta}~=~-y_{[\alpha}\bar y_{\beta]}~-~-X_{[\alpha}y_{\beta]}, $$ so the Weyl tensor is composed as $$ \begin{align}C_{\alpha\beta\mu\nu}&~=~ \Psi_0U_{\alpha\beta}U_{\mu\nu} \\ &\, \, \, ~+~\Psi_1(U_{\alpha\beta}W_{\mu\nu}~+~W_{\alpha\beta}U_{\mu\nu}) \\ &\, \, \, ~+~\Psi_2(V_{\alpha\beta}U_{\mu\nu}~+~U_{\alpha\beta}V_{\mu\nu}~+~W_{\alpha\beta}W_{\mu\nu}) \\ &\, \, \, ~+~\Psi_3(V_{\alpha\beta}W_{\mu\nu}~+~W_{\alpha\beta}V_{\mu\nu}) \\ &\, \, \, ~+~\Psi_4V_{\alpha\beta}V_{\mu\nu}\end{align}, $$ for $\Psi_i$ Weyl scalars. These define different physics; $\Psi_2$ gives the vacuum around a central source, such as a black hole, $\Psi_4$ are transverse modes and $\Psi_1,~\Psi_3$ are in and out directed longitudinal modes.

Each of the bivectors may be expressed according to a vierbein $U^a~=~E^\alpha u^a_\alpha$, where now the Latin indices refer to spacetime and Greek indices correspond to an internal space given by the root vectors of a Lie algebra. We can then see that $U_{ab}~=$ $2[E^\alpha,~E^\beta]U_beta^{[b}U_\alpha^{a]}$. The Weyl tensor is then $$ C^{abcd}~=~ 2\Psi_0[E^\alpha,~E^\beta][E^{\alpha'},~E^{\beta'}]U^{ab}_{\alpha\beta}U^{cd}_{\alpha'\beta'} $$ $$ ~+~2\Psi_1[E^\alpha,~E^\beta][E^{\alpha'},~E^{\beta'}](U^{ab}_{\alpha\beta}W^{cd}_{\alpha'\beta'}~+~W^{ab}_{\alpha\beta}U^{cd}_{\alpha'\beta'}) $$ $$ +~2\Psi_2[E^\alpha,~E^\beta][E^{\alpha'},~E^{\beta'}](V^{ab}_{\alpha\beta}U^{cd}_{\alpha'\beta'}~+~U^{ab}_{\alpha\beta}V^{cd}_{\alpha'\beta'}~+~W^{ab}_{\alpha\beta}W^{cd}_{\alpha'\beta'}) $$ $$ +~2\Psi_3[E^\alpha,~E^\beta][E^{\alpha'},~E^{\beta'}](V^{ab}_{\alpha\beta}W^{cd}_{\alpha'\beta'}~+~W^{ab}_{\alpha\beta}V^{cd}_{\alpha'\beta'}) $$ $$ +~2\Psi_4[E^\alpha,~E^\beta][E^{\alpha'},~E^{\beta'}]V^{ab}_{\alpha\beta}V^{cd}_{\alpha'\beta'} $$

where the bi-vierbeins $U^{ab}_{\alpha\beta}$ are evidently defined. The nature of the gauge field or gauge-like field associated with these Lie algebraic roots is discussed in the last section. \vskip.1in The root vectors $E^\alpha$ obey the commutators $$ [E^\alpha,~E^\beta]~=~N^{\alpha\beta}H^{\alpha+\beta}, $$ where $H^{\alpha+\beta}$ are the weights. With any Lie algebra there are elements that are analogous to $a$ $a^\dagger$ for the harmonic oscillator, which are the standard roots $E^\alpha$ and $a^\dagger a$ that correspond to the weights $H^\alpha$. For the gauge theoretic description of the $\hat C$ operators the result is linear in the weight. For gravitation the operator is quadratic in the weights. The Weyl tensor for type $D$ and $II$ solutions are eigenvaled with $C_{abcd}U^bU^c~=~\lambda U^aU^d$. It is possible to see the Weyl tensor for these eigenvalued Petrov types obeys $$ C_{abcd}U^bU^c~=~\lambda E_\alpha E_\delta U^\alpha_a U^\delta_d. $$ Consequently the Weyl tensor for these eigenvalued Petrov types obeys $$ C_{abcd}U^bU^c~=~N_{\alpha\beta}N_{\gamma+\delta}H_{\alpha+\beta}H_{\gamma+\delta}U^\alpha_a U^\beta_bU^\gamma_cU^\delta_dU^bU^c~ $$ $$ =~N_{\alpha\beta}N_{\gamma+\delta}H_{\alpha+\beta}H_{\gamma+\delta}E^\beta E^\gamma U^\alpha_a U^\delta_d~=~\lambda E_\alpha E_\delta U^\alpha_a U^\delta_d. $$ The commutators of the Lie algebraic roots concern the observables $A_i$ and $B_i$ on either side of the apparatus. In the case of a Lie algebra a gauge transformation, or the introduction of a force, transforms these operators relative to each other. This results then in a modification of the Tsirelson bound. Similarly, for gravitation the parallel translation of these operators on either side of the apparatus does the same.

This is a bit formal, We may thought think of the weights as the states of gluons. These are formed from the commutator of the root vectors. The product of weights sum over the internal index of the Lie algebra and form a Weyl tensor. In this way we have derived some aspects of a graviton from the group representation of a gauge particle. In particular the graviton is a product of the gauge particle.


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