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As I understand it, the scalar curvature is a function that assigns a real number between $]-\infty,\infty[$ to each point $(x,y,z,t)$ of a manifold:

$$ R:\mathbb{R}^4\to \mathbb{R} $$

I am having difficulty picturing the scalar curvature and why it is treated as an independent quantity. Specifically, according to Wikipedia "To each point on a Riemannian manifold, it assigns a single real number determined by the intrinsic geometry of the manifold near that point.". Now, I am confused. What does $R$ being determined means in the case. Does it mean I can use the metric, or some other quantity, to solve for the scalar curvature? If this is the case, why is the Hilbert-Einstein action contain both $R$ and $g$ if $R$ is solved from $g$? If this is not the case, then the Wikipedia explanation would be inside-out; would it rather be the scalar curvature that defines the intrinsic geometric of the manifold near that point?

For a given $g$, can $R$ be any choice of arbitrary, possibly smooth, function - each choice of $R$ defining a different manifold?

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    $\begingroup$ Specifically in GR which uses a Levi-Civita connection does the metric not determine the connection coefficients and by extension the Riemann tensor $R_{\mu\nu\alpha\beta}$ whose full contraction $g^{\mu\alpha}g^{\nu\beta}R_{\mu\nu\alpha\beta}=R$? Does this help? $\endgroup$
    – Charlie
    Nov 16 '20 at 14:20
  • $\begingroup$ @Charlie Can one prescribe both the Ricci scalar and the metric, and obtain a valid GR solution? From your comment, it looks like if I were to prescribe both the Ricci scalar and the metric, then it would imply a specific Riemann tensor (based on the contraction relation) and thus it would work out. I am trying to understand if $R$ and $g$ can be understood as two independently prescribable "inputs" of the Hilbert-Einstein action, for any smooth function and metric, respectively. $\endgroup$
    – Anon21
    Nov 16 '20 at 14:26
  • $\begingroup$ Someone more experienced may correct me, but it is my understanding that a metric field which solves the EFE uniquely determines the geometry of spacetime, which includes the curvature tensors and the Ricci scalar. Otherwise the metric isn't much of a "solution" to the EFE if one also has to prescribe a scalar $R$ field. $\endgroup$
    – Charlie
    Nov 16 '20 at 15:07
  • $\begingroup$ Wikipedia seems to verify this: en.wikipedia.org/wiki/…. "The metric $g$ completely determines the curvature of spacetime." $\endgroup$
    – Charlie
    Nov 16 '20 at 15:08
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In Riemannian geometry, there are in principle two fundamental quantities: the metric $g$ and the connection $\nabla$. Each one may be chosen independently of the other, or even be left undefined (that is, you can have only the metric, or only the connection).

The Riemann curvature tensor $R^a{}_{bcd}$ depends only on the connection, as can be seen from its expression in a coordinate chart:

$$R^a{}_{bcd} = \partial_c \Gamma^a{}_{db} + \Gamma^a{}_{ce} \Gamma^e{}_{db} - (c \leftrightarrow d).$$

The Ricci tensor can be defined directly from this as $R_{bd} = R^a{}_{bad}$. If you also have a metric, you can define the curvature scalar as $R = g^{bd} R_{bd}$.

You can see that in general, the curvature scalar depends on both the metric and the connection. If you have one of them (say, the metric) and you want to choose $R$ as an independent quantity, you'll have to set up equations and see if you can choose the other one (the connection if you fixed the metric) so that $R$ equals your desired function; this may or may not be possible in general, I don't know.

This is the general situation, but in general relativity the connection is not independent; we use the Levi-Civita connection, which depends only on the metric, so that all the other quantities also depend only on the metric. And the same logic applies: if you want $R$ to be some specific function, you might be able to choose a metric that gives you your desired $R$, but it's not straightforward.

And to answer your last paragraph: the EH action (or the field equations) contain $R$, but it is understood that it is really a function of the metric, which we don't write out explicitly because, well, it's just $R$.

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  • $\begingroup$ Does this carry any weight in the case of GR en.wikipedia.org/wiki/Prescribed_scalar_curvature_problem, or does it require full flexibility of the connection such that it is not necessarily the Levi-Civita connection? $\endgroup$
    – Anon21
    Nov 16 '20 at 17:07
  • $\begingroup$ @Alexandre I think that problem assumes the Levi-Civita connection, but it's not very relevant for GR because it applies to a Riemannian manifold. I'm not sure about the semi-Riemannian case. $\endgroup$
    – Javier
    Nov 16 '20 at 17:16
  • $\begingroup$ Right! What about this guy arxiv.org/pdf/math/0409435.pdf? He seems to claim the prescribed scalar curvature problem holds for pseudo-Riemannian metrics. $\endgroup$
    – Anon21
    Nov 16 '20 at 17:23
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    $\begingroup$ @AlexandreH.Tremblay I'm a bit skeptical since the talk slides state a slightly different theorem than the arxiv paper (in the 4-dimensional case, the latter requires the prescribed function to be somewhere positive while the former does not). In any event, these results are claimed only for compact manifolds, which typically are not of much interest as spacetimes since any compact manifold with Lorentzian metric has closed timelike curves, i.e. has causal structure often thought of as unphysical. $\endgroup$
    – jawheele
    Nov 17 '20 at 3:50
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    $\begingroup$ @AlexandreH.Tremblay The arxiv paper result also requires a nonempty boundary for n=4, whereas the slides claim not to. If the result does require a nonempty boundary, that's another reason it's less compelling for the GR use case, since we typically don't think of our spacetimes as including a boundary due to a lack of physical evidence pointing to one. $\endgroup$
    – jawheele
    Nov 17 '20 at 3:59
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The curvature tells you something about the local structure through differential geometry. That does not fully constrain the topology of the manifold, however, which is a global property.

A simple example is that the (2D) surface of a cylinder has $R=0$ everywhere, the same as the open plane. Locally they look the same from the perspective of curvature, but globally there's a clear difference of topology since one direction is compact on the cylinder.

Now it seems that you asked a couple of different questions in the title and in the text of your original post. Trying to answer a few of them:

  1. Knowing $R$ everywhere is not enough to tell you the global structure of the manifold.
  2. It's also generically true that a Riemannian manifold is typically defined as a pair $(M,g)$ of a manifold $M$ and a metric $g$. The Ricci scalar $R$ is defined in terms of $g$ (but not explicitly $M$).
  3. The Ricci tensor is defined in terms of $g$ and its derivatives. Having both $R$ and $g$ in the Einstein-Hilbert action is just a compact way of representing how the derivatives of $g$ enter the action. It's not fundamentally different than any other action that depends on a field and its derivatives. The result of varying the action in such a case typically gives you a set partial differentials equation to solve for the fields - In this case the Einstein equation.
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In general, you seem very focused on the scalar curvature. The scalar curvature is just one measure of curvature. A spacetime can be curved and have zero scalar curvature, as in, e.g., a gravitational wave.

For a given g, can R be any choice of arbitrary, possibly smooth, function...

No. R can be determined from g.

...each choice of R defining a different manifold?

The manifold is a topological object. Changing the metric doesn't change the manifold.

Can one prescribe both the Ricci scalar and the metric...

No. R would have to be consistent with g. It adds no information and would have to be chosen to be consistent with the metric.

and obtain a valid GR solution?

Any metric is a solution to the field equations, unless something is specified about the stress-energy.

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