It is well-known that the equation for stationary solutions of the Einstein-Hilbert functional (without matter and cosmological constant, which is irrelevant here):
$$S = \int_M R \mu_g,$$ is given by the Einstein's field equation: $$Ric -\frac{1}{2}R g = 0, $$ where $\mu_g$ is the canonical volume form given by the metric $g$, $Ric$ is the Ricci curvature and $R$ is the Ricci scalar.

The standard derivation of the above statement seems to be a not so hard but not so pleasant direct calculation, either in coordinates or abstract indices, expanding everything in terms of the Christoffel symbol and eventually in terms of $g$ and then calculus.

My questions is: is there a more geometric and coordinate-free way to derive this?


1 Answer 1


This can be found in Besse "Einstein Manifolds", in chapter 4.

The idea is to use Koszul formula for the Levi-Civitta connection to compute the derivative of the curvature with respect to the metric. Bianchi identities also help.

  • $\begingroup$ Thank you very much. That is indeed a coordinate-free calculation. Still, it is just the standard derivation in the coordinate-free notation. So I didn't formulate my problem well as I was hoping for something more geometrical rather than algebra, say in terms of parallel transport or something of that kind. Thanks all the same. $\endgroup$
    – Lizao Li
    Sep 10, 2012 at 20:54
  • 1
    $\begingroup$ @Thomas Richard Is there some free resource in internet on this argument? $\endgroup$
    – asv
    Mar 20, 2018 at 16:46

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