# Is it possible to write the Hilbert Action as a product of two identical tensors?

We know the Maxwell action can be written as the tensor product of the tensor $$F^{ab}$$ with itself. $$F^{ab}F_{ab}$$

[Edit: This bit I forgot to mention in the original quesiton] Using the product rule one can write the action $$\int\sqrt{-g}Rdx^4$$ in terms of first derivatives only (ignoring boundary terms). Call this Lagarangian $$B$$. So:

$$B=\sqrt{-g}\left( g^{ab}g^{de}g^{cf} +2 g^{ac}g^{bf}g^{de} + 3g^{ad}g^{be}g^{cf} -6 g^{ad}g^{bf}g^{ce} \right)\partial_c g_{ab}\partial_f g_{de}$$

Simimlarly is there a tensor (or indeed a non-tensor matrix object) $$P^{abc}$$ such that $$P^{abc}P_{abc}=B$$? $$P$$ should contain only first derivatives of $$g$$. i.e. ignoring boundary terms: $$\int \sqrt{-g}R dx^4 = \int\sqrt{-g}P^{abc}P_{abc}dx^4$$

Or is there a simple proof that this is not possible?

My guess is you would write:

$$P_{abc} = \alpha_1 \partial_a g_{bc} + \alpha_2 \partial_b g_{ac} + \alpha_3 \partial_c g_{ab} + \alpha_4 g_{ab} g^{ef}\partial_c g_{ef} + \alpha_5 g_{ac} g^{ef}\partial_b g_{ef}+ \alpha_6 g_{bc} g^{ef}\partial_a g_{ef}$$

And see what values of $$\alpha$$ would possibly solve it. Even if the $$\alpha$$ are non-commutative.

Edit: Using computer software I think it can be done where the $$\alpha$$ are complex numbers. Don't think there is a real valued solution.

• You are inconsistent about whether $P$ has two indices or three. – G. Smith Apr 21 at 17:55
• It’s rude to invalidate answers by changing the question after an answer is written. – G. Smith Apr 21 at 17:56
• Yes, I made a mistake because a derivative of the metric would have 3 indices. – zooby Apr 21 at 17:56
• @Smith. Sorry it was a mistake I made in the question. Not trying to be rude. – zooby Apr 21 at 17:57
• You should write your change as an addendum and make it clear that it was added after my answer. Then someone else can answer your revised question. – G. Smith Apr 21 at 17:58

Though not precisely what you intended, you might be interested in MacDowell–Mansouri formulation of gravity.

This formalism combines the Levi-Civita connection and coframe field into a single physical field, leading to a gauge theory Lagrangian: $$S_\text{MM}[A]=-\frac{1}{2\Lambda} \int \mathrm{tr}\,(\hat{F}\wedge \star \hat{F}),$$ with either de Sitter or anti-de Sitter group as the gauge group (depending on the sign of cosmological constant).

This action is classically equivalent to Einstein–Hilbert action with cosmological constant (the difference between these actions is proportional to purely topological Gauss–Bonnet term that does not change the EFE's).

The original paper:

• MacDowell, S. W. & Mansouri, F. (1977). Unified geometric theory of gravity and supergravity. Phys. Rev. Lett. 38 (14): 739–742. doi:10.1103/PhysRevLett.38.739.

More accessible exposition could be found in thesis by D. Wise, see also his paper or these slides.

• Thanks. The math is a little complicated for me. Could you elaborate? I only understand the tensor formalism not the wedge, hodge star, hat formalism. – zooby Apr 21 at 21:08
• These are tensors “under the hood”. You can think of this notation as a consistent way to hide indices. For example, $\hat{F}$ is an (a)dS algebra-valued 2-form, which means that underneath it has 2 internal indices and 2 indices of covariant antisymmetric tensor. The slides I linked to give the most concise definitions of these objects (also I noticed that some coefficients like $1/\ell$ are omitted there). – A.V.S. Apr 22 at 6:12

I found one answer by using computer algebra software. Unfortunately the answer has complex coefficients:

$$P_{abc} = \frac{1}{2}\left(3−\sqrt{-3}\right) \partial_a g_{bc} +\frac{1}{2}\left(-\sqrt{-3}-3\right)\partial_b g_{ac}\\ +(0.3735i+0.8771)g_{bc}g^{ef} \partial_a g_{ef} +(0.86146i−0.2620)g_{bc}g^{ef} \partial_a g_{ef}$$

Looks like there is an infinite number of solutions in fact if we let the coefficients be complex numbers. And zero solutions if the coefficients must be real.

Since it only can be solved in complex numbers I don't think it is of any significance. Unlike for Maxwell which has the simple $$F_{ab}=\partial_a A_b-\partial_b A_a$$

A better solution might be just subtitute in $$\partial_a g_{bc} = \Gamma_{bac}-\Gamma_{cab}$$ in the equation for $$B$$ to get a kind of square term involving just the Christoffel symbols.