# (Basic) question regarding Einstein-Hilbert-functional / total scalar curvature

I got a question regarding the total scalar curvature / Einstein-Hilbert-functional.

I found it in Peter Topping's book "Lectures on the Ricci flow" (you can download it for free here: http://homepages.warwick.ac.uk/~maseq/RFnotes.html ) at the very beginning of chapter 6.

I'm fine with the calculations he does on p. 67. Then in the next page he defines the Gradient of E, which looks... pretty random to me and so the following question arose:

Is it possible to make the domain of E (i.e. the set of smooth Riemannian metrics on a closed Riemannian manifold) into a Riemannian manifold, then look at E as a (differentiable?) function on this manifold and show that its gradient is equal to the definition that Peter Topping gave?

Thanks in advance for any help!

• Welcome to MathSE and thank you for you question. I admit that it is not easy to find that a very similar question has been already asked and answered here. Please look at this one. – Yuri Vyatkin May 11 '13 at 9:52
• Thank you, that provided all the information I wanted! I really appreciate your help. – zybrr May 11 '13 at 10:13
• Thinking of it, there is one more thing that I want to understand. In the link you provided, smiley06 states that "in this case it is L^2". I only know the space L^2 from integration theory (= square integrable functions on a measure space). So, do you simply define L^2:={ Riemannian metrics on a fixed manifold} and show that it's a Hilbert space with the inner product given in the link, or is there more behind it? – zybrr May 11 '13 at 11:23
• I guess that @smiley06 speaks about L^2:={symmetric tensors on a fixed manifold with the L^2 inner product} that is easily shown to be a Hilbert space. I am not too sure if it makes sense to deal with the manifold of all Riemannian metrics but you may ask a question on that on mathoverflow.net where P.Michor is frequently seen :-) – Yuri Vyatkin May 12 '13 at 11:02
• By the way, please look at this my answer too. – Yuri Vyatkin May 13 '13 at 8:25