Given a manifold with a metric (M,g) you can construct the Ricci scalar R via \(R_{ab} = R^{c}_{acb}\) and \(R = g^{ab}R_{ab}\). Suppose I've a second space (M',g') with Ricci scalar R'. I can make a combination of these to define a space with a matric \((M\times M',g \oplus g')\). Is it the case that the resultantr Ricci scalar is just R+R'? The alternate construction (which is actually what I'm considering) is to have a space split, \(S \to S_{1} \times S_{2}\) such that the metric of S splits into block diagaonal form : \(G \to \left( \begin{array}{cc} g_{1} & 0 \\ 0 & g_{2}\end{array}\right)\) This is such that \(g_{1}\) doesn't depend on the coordinates of \(S_{2}\) and vice versa for \(g_{2}\). I'm pretty sure this results in the Riemann curvature and its contractions down to R being of the same kind of block diagonal form (though its rank 4 so the splitting is the rank 4 version of the metric splitting) since there's no mixing at all. I just want someone to confirm this before I start doing what could be reams of pointless algebra. I get the feeling its either obviously true or obviously false so I am probably just being thick which ever way it goes. Thanks.
Hi AN. The consensus in the office is that the Ricci scalars are simply summed. That makes sense because the Riemann tensor will be "diagonal," Ie elements from S_1 and S_2 will not mix. When you make your contractions with the metric there is nothing to mix the components and you'll end up with a simple sum.
Let \(x_i\in S_i\). Then if \(g_i=g_i(x_i)\) then the statement is true. But if \(g_i=g_i(x_1,x_2)\) then the (connection and so) curvature will have mixed components.
That sounds like I expected. However there is the issue of the fact the metric doesn't split additively but multiplicatively. \(G = g_{1} \oplus g_{2}\) then \(|G| = |g_{1}||g_{2}|\) and \(R(G) = R_{1}+R_{2}\) and so the E-H action goes as \(\int \sqrt{|G|}R(G) = \int \sqrt{|g_{1}|}\sqrt{|g_{2}|}(R_{1}+R_{2})\) If you assume that \(g_{2}\) is associated to a finite volume region then by using \(\int_{M} d^{N}x \to \int_{M_{1}}d^{n_{1}}x \int_{M_{2}}d^{n_{2}}y\) and can integrate over \(M_{2}\) for the \(R_{1}\) dependent term using \(\int_{M_{2}} \sqrt{|g_{2}|} d^{n_{2}}y = Vol(M_{2})\) so \(\int_{M} d^{N}x \,\sqrt{|G|}R = Vol(M_{2}) \int_{M_{1}}\sqrt{|g_{1}|}R_{1} + \left( \int_{M_{1}} d^{n_{1}}x \sqrt{|g_{1}|} \right)\left( \int_{M_{2}} d^{n_{2}}y \sqrt{|g_{2}|}R_{2} \right)\) The fact part of the action becomes dependent on the volume of the finite (or compact) space is something seen in the usual Kaluza-Klein compactifications in string theory. The other term isn't so familiar. Is it natural not to expect the E-H action to split into two parts in the way the space-time does? I was expecting that but the algebra doesn't lead to it. In fact the \(M_{1}\) dependent part of the second term would be the volume of \(M_{1}\) if the space is finite but it isn't finite so I'm wondering if that's even a valid expression. Obviously if \(R_{2}=0\) then you reduce to an expression which is finite and splits into an \(M_{1}\) and an \(M_{2}\) part and it just so happens that's a common thing in string theory as Calabi-Yaus are \(R_{ab} = 0 = R\) flat but that's just a particular case. Anyone know if that non-splitting of the E-H action is to be expected and is well defined? The E-H action reproduces the Einstein field equations under variation of g and contains kinetic terms for the metric, unlike \(\int_{M_{1}} \sqrt{g}\) which doesn't have any derivatives and doesn't seem to be a good action term for infinite \(M_{1}\). I must admit its been a while since I worked on the explicitly GR formulation of curved spaces so I might be asking some real clangers here.
I think I see the issue; you could make \(g_2\) depend on \(x_1\in M_1\) but then it will ruin the curvature decomposition, or maybe some kind of "renormalization" trick. But would it really matter if the action is bounded or not? If you are interested in the differential equations the action has to offer, those are derived from compactly supported variations so the "unboundedness" would not cause any trouble. I put unboundedness in quotes because to derive the differential equations, you will be in essence considering "local actions" that come from integrating over small regions of \(M_1\) instead of the whole of it.
Is the Ricci scalar really the sum of the corresponding Ricci scalars from the corresponding Riemannian manifolds? Take a cylinder: \( M= \mathbf{R} \times S^1 \) Then we know this has zero scalar curvature. But the curvature of the real line is vanishing, and the curvature of circle is positive. Have I misunderstood?
I have 2 questions here: What do you mean by the curvature of a 1 dimensional object, and are you sure that the curvature of a circle is non zero? I tend to think of the \(S^1\) as being equivalent to \(\mathbb{R}^1\), which is flat.
Ok, every manifold whose dimension is less than 2 is conformally flat. Does that mean that the manifold is Ricci flat (and therefore has zero Ricci scalar)? I can't think of any other outcome from computing the Riemann tensor than zero. That could be too naive a method for a single dimensional space though.
Oh dear - not a good start to the day. The curvature I was refering to was the standard curvature of a plane curve, which has little to do with the point in question!
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Yes, normally you always preface anything like this with phrases like "Assuming compact support" or "The fields decay sufficiently at infinity" so that everything is well defined. I've never really put much thought to it, though I have seen books discussing the E-H action and comment that \(\int \sqrt{|G|}\) isn't the right action and then the next ansatz is \(\int \sqrt{|G|}R\) which gives the right field equations. So if you're going at it from the point of view of wanting the right equations you just ignore \(\sqrt{|G|}\) and use \(\sqrt{|G|}R\). If you then consider the simplest space-times of dS, Minkowski and AdS you get that \(\int sqrt{|G|}R\) is not always finite with them. I suppose you're not really supposed to ask what value the action takes but only consider how variations in the fields might vary the action.
