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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.

Qmechanic
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  • You are inconsistent about whether $P$ has two indices or three. – G. Smith Apr 21 '20 at 17:55
  • It’s rude to invalidate answers by changing the question after an answer is written. – G. Smith Apr 21 '20 at 17:56
  • Yes, I made a mistake because a derivative of the metric would have 3 indices. –  Apr 21 '20 at 17:56
  • @Smith. Sorry it was a mistake I made in the question. Not trying to be rude. –  Apr 21 '20 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 '20 at 17:58
  • I don’t consider that edit satisfactory. I am going to delete my answer. I will be hesitant to answer any further questions from you. Best wishes on getting answers from others. – G. Smith Apr 21 '20 at 18:03
  • Somewhat related question: https://physics.stackexchange.com/questions/340371/why-does-the-analogy-between-electromagnetism-and-general-relativity-differ-if-y – knzhou Apr 21 '20 at 18:09
  • @knzhou Similar yes. Basically I was watching Weinstein's lecture on his "geometric unity" theory, and this made me think of this question. But the difference is I'm writing it in terms of first derivatives only. So it might be easier. –  Apr 21 '20 at 18:18

1 Answers1

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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.

A.V.S.
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  • 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. –  Apr 21 '20 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 '20 at 06:12