Visualizing Ricci scalar curvature

I am trying to learn more about Ricci scalar curvature. I am trying to get an image in my head of what scalar curvature actually represents about the curvature of a manifold. The most familiar image I have of a scalar giving me information about a space is a scalar field, like temperature. Is that the right way to think about scalar curvature? If not, how should one envision scalar curvature?

According to Wikipedia here, scalar curvature can be characterized as a multiple of the average sectional curvatures at a point.

• Perhaps something like this might be useful although it comes from the world of physics and GR , not the Ricci flow of Hamilton but it might be of interest. math.ucr.edu/home/baez/gr/ricci.weyl.html – Alan Jan 29 '15 at 0:04
• @Alan That was cool. Question: why does the Ricci curvature tensor measure changes in the volume of a ball? Suppose we don't specify that our manifold has the structure of a ball. How can we know the Ricci tensor is applicable to that manifold? That is, is Ricci limited only to manifolds with the structure of a ball? – Stan Shunpike Jan 29 '15 at 0:13
• What do you mean by structure of a ball - since we are talking about manifolds every point has neighbourhood which is a ball. And I should add that the curvature is defined locally - we assign the curvature at each point. – Joseph Zambrano Jan 29 '15 at 1:29
• @JosephZambrano Where does it say that in the definition? I thought a smooth manifold was a topological manifold with a smooth structure. and that the only characteristics of a topological manifold are that its Hausdorff, second-countable, and with nbhd at each point homeomorphic to $\Bbb{R}^n$. It never usually says anything about a ball... – Stan Shunpike Jan 29 '15 at 1:34
• A ball is homeomorphic (diffeomorphic) to $\mathbb{R}^n$. – Joseph Zambrano Jan 29 '15 at 1:35

Some of this parallel remains in higher dimensions, since the sign of curvature determines the asymptotic volume comparison for geodesic balls (positive curvature $\implies$ less-than -Euclidean volume, negative $\implies$ greater than Euclidean). However this is on small scale only, and therein lies a problem: the scalar curvature does not deliver much global geometric information.
• The set of all points at distance less than $r$ from the center. The distance on a Riemannian manifold is the length of shortest geodesic, hence the name. – user147263 Jan 29 '15 at 2:03
• @Fundamental: Actually, in dimensions $>2$ there are no obstructions for a manifold to have a metric of negative Ricci curvature. For positive scalar curvature, of course, there are many restrictions and none of them simple indeed... – Moishe Kohan Jan 30 '15 at 1:08