# Work on an Einstein-Hilbert type action but with the *absolute value* of scalar curvature?

This is only my second question on mathoverflow, so my apologies if this would be more appropriate at a physics site. My question concerns a modification to the Einstein-Hilbert action. The standard action is given (in the absence of matter and with cosmological constant $\Lambda=0$) by

$$\mathcal{S_{EH}}(g_{\mu\nu}) = \int_M R \sqrt{-g}\mbox{ }d^4x$$

where $M$ is a (compact) differentiable 4-manifold, $g_{\mu\nu}$ is a Lorentzian metric on $M$, $R$ is scalar curvature and $\sqrt{-g}\mbox{ }d^4x$ is the standard volume form. Critical points of this action (with respect to variations in $g_{\mu\nu}$) give Lorentzian metrics which are solutions to Einstein's field equations for general relativity.

QUESTION: Does anyone know of work using a similar action, but where the absolute value $|R|$ appears instead of $R$? That is, I'm interested in references to previous work concerning the action $$\mathcal{S}(g_{\mu\nu}) = \int_M |R| \sqrt{-g}\mbox{ }d^4x.$$

Given the huge amount of interest in quantum gravity, I would assume that someone has examined this. However, I was unsuccessful in my searches. I'm not a physicist, so perhaps I'm missing some bit of terminology that is standard. Any help pointing me in the right direction would be greatly appreciated!

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In the vacuum case this is not greatly different from the Einstein-Hilbert action.

Let $(M,g)$ be a classical solution to the variational problem as you posed. Suppose $p\in M$ is such that $R(p) \neq 0$, then by continuity in a small neighborhood of $p$, the scalar curvature $R$ is signed, and hence locally in that neighborhood it is also a critical point to the Einstein-Hilbert action. But then it must be Ricci flat, contradicting the assumption that $R \neq 0$ at $p$.

Conversely, if $(M,g)$ is a classical solution to the Einstein-Hilbert variational problem, then it is Ricci flat and hence scalar flat. And hence you have that all Einstein-vacuum solutions are also solutions to the critical point problem you posed.

Going back forwards again, note that by definition any scalar flat 4 manifold will be a minimizer of the action. Hence you have that for the vacuum problem of your proposed action:

There are no critical points which do not minimise the action; the action minimisers are precisely the scalar flat Lorentzian 4 manifolds.

In any case, if you really are interested in this action, for literature searches the relevant keyword is f(R) gravity theories.

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@Willie Wong: Thanks, Willie! Great argument ... and exactly what I was looking for! – Aaron Trout Jul 26 '12 at 18:10
Classically the theories are the same, but quantum mechanically, think path integral, they will differ. – Kelly Davis Jul 26 '12 at 21:54
@Kelly: as I will not pretend to know how to think quantum mechanically about gravity, I cannot comment on that. :-) – Willie Wong Jul 26 '12 at 22:12