# 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!

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.

• @Willie Wong: Thanks, Willie! Great argument ... and exactly what I was looking for! Jul 26, 2012 at 18:10
• Classically the theories are the same, but quantum mechanically, think path integral, they will differ. Jul 26, 2012 at 21:54
• @Kelly: as I will not pretend to know how to think quantum mechanically about gravity, I cannot comment on that. :-) Jul 26, 2012 at 22:12