Is there a "square root" version of the Einstein field equation?

Question

It is well known that the Klein-Gordon equation have a kind of "square root" version : the Dirac equation.

The Maxwell equations can also be formulated in a Dirac way.

It is also well known that the metric of general relativity have a kind of "square root" version : the tetrad field (or vierbein) of components $e_{\mu}^a(x)$ : \begin{equation}\tag{1} g_{\mu \nu}(x) = \eta_{ab} \, e_{\mu}^a(x) \, e_{\nu}^b(x). \end{equation} Now, a natural question to ask is if the full Einstein equations : \begin{equation}\tag{2} G_{\mu \nu} + \Lambda \, g_{\mu \nu} = -\, \kappa \, T_{\mu \nu}, \end{equation} could be reformulated for the tetrad field only (or other variables ?), as a kind of a "Dirac version" of it ? In other words : is there a "square root" version of equation (2) ?

Answer 1

1. Since Nature has fermionic matter we are anyway ultimately forced to rewrite the metric in GR in terms of a vielbein (and introduce a spin connection). See e.g. my Phys answer here. The fermionic matter obeys a Dirac equation in curved spacetime. This however would not amount to a square root of EFE.

2. There exist supersymmetric extensions of GR, such as, SUGRA.

3. Another idea is to consider YM-type theories as a square root of GR, or GR as a double copy of YM. See e.g. the Ashtekar formulation or the KLT relations.

Answer 2

By taking the "Dirac square root" of the Hamiltonian constraint for GR, you naturally end up with Supergravity...so in some appropriate sense, SUGRA "is" a "square root" of GR. For more on this, see:

• Romualdo Tabensky, Claudio Teitelboim, "The square root of general relativity". Physics Letters B 69 no.4 (1977) pp 453-456. Eprint
• Claudio Teitelboim, "Supergravity and Square Roots of Constraints". Phys. Rev. Lett. 38 (1977) 1106. Eprint