# What is the intuitive meaning of the scalar curvature R?

Background:

Let $M$ be a smooth, Riemannian manifold with metric $g$ and dimension $n$. Let $R^a_{bcd}$ be the Riemann tensor with respect to the Levi-Civita connection for $g$.

Question:

Is there any rigorous result that gives a good intuitive sense of the meaning of the scalar curvature $R = R^{ab} R_{ab}$?

Discussion:

What I have in mind is something like the following:

For $n =2$, the volume of a geodesic ball of radius $\epsilon$ in $M$ is $\pi \epsilon^2 [1 - (R / 48) \epsilon ^2 + O(\epsilon^4) ]$. I may have the numerical factors wrong, but the point is this: $R$ tells you the difference between the volume of a geodesic ball and an ordinary Euclidean ball (for small radius). That's the kind of result I'm looking for.

My problem with this result is that it holds in normal geodesic coordinates but not in general coordinates. (Note that a choice of coordinates is necessary to define 'a geodesic ball of radius $\epsilon$'). If you know how to generalize this result to arbitrary coordinates, or you know another result that gives some intuition for $R$, please let me know.

Thanks for any help!

-
Have you looked up comparison theorems, such as Bishop-Gromov or Heintze-Karcher? Also, coordinates are not necessary to define geodesic $\epsilon$-balls; using geodesics to define distance induces a metric space structure on a Riemannian manifold. –  Neal Dec 12 '11 at 2:25
@Neal: I think that Bishop-Gromov only applies when you have a bound on Ricci curvature, which is a significantly stronger condition than having a bound on scalar curvature. Likewise, Heintze-Karcher needs sectional curvature bounds, as I understand it (but I'm much less familiar with this). –  Jason DeVito Dec 12 '11 at 2:34
I wasn't clear. My idea is that those theorems might help form intuition about the relationship between curvature (but not necessarily scalar) and other geometric quantities, such as the volume of distance balls. –  Neal Dec 12 '11 at 2:58

Let us start with two familiar settings. On the sphere of radius 1, a spherical cap of geodesic radius $\rho$ has area $2 \pi (1 - \cos \rho).$ In the hyperbolic plane of curvature $-1,$ the area of a disk of radius $\rho$ has area $2 \pi ( \cosh \rho - 1).$ Writing out just a few terms of the power series for $\cos$ and $\cosh$ tells you the comparison to $\pi \rho^2.$

Actually, it is not necessary to specify a coordinate system of any sort to define distance on a Riemannian manifold, that is what geodesics are for. As a result, for small radii, the geodesic ball is also well-defined. Furthermore, in a pleasant feature, for small enough radii the geodesics do meet the sphere bounding the ball orthogonally, this is usually called the Gauss Lemma

Other than that, see Paul's answer...

-
Thanks for your reply. I should have mentioned that I ultimately want to do the same thing for a semi-Riemannian manifold. I understand how (in the Riemannian case) one can use geodesics to define a metric on the manifold, and then use the same definition of 'ball' as is used for any metric space. I wonder if you might take a look at the comment I posted below Paul's answer, in which I try to give a definition of 'geodesic ball' that also works in the semi-Riemannian case. I think something is wrong with my definition -- I'd appreciate any help. –  marlow Dec 19 '11 at 1:22
@marlow You still get geodesics, but I think you had better find out what actually is done. See books.google.com/books/about/… or amazon.com/… –  Will Jagy Dec 19 '11 at 3:42

What you have mentioned for $n=2$ is in fact true for higher dimension too: Quoted from here, we know that the ratio of the $n$-dimensional volume of a ball of radius $\epsilon$ in the manifold $M$ to that of a corresponding ball in Euclidean space $\mathbb{R}^n$ is given by, for small $\epsilon$, $$\frac{\mbox{Vol}(B_\epsilon(p)\subset M)}{\mbox{Vol}(B_\epsilon(p)\subset\mathbb{R}^n)}=1-\frac{R}{6(n+2)}\epsilon^2+O(\epsilon^4).$$ (Therefore, you are right with the constant for the case when $n=2$).

-
I have been trying to prove this in the n=2 case, without success. Oops - I posted this comment before I finished typing. More is coming. $\times$ –  marlow Dec 19 '11 at 1:03
I am having trouble proving this in n=2. We want to find the volume of the geodesic ball of radius $\epsilon$ centered at $p$. Every source I've seen has a 'geodesic ball' as the set of all points $\gamma(1) \in M$, where $\gamma$ is any geodesic satisfying $\gamma'(0) = v$ for some $v\in T_pM$ s.t. $<v,v> \lt \epsilon$. Would it possible to reverse E and 1, considering instead the set of all points $\gamma(\epsilon) \in M$, where $\gamma$ is any geodesic satisfying $\gamma'(0) = v$ for some $v\in T_pM$ for which $<v,v><1$? I tried the latter in a special case but it appears not to work. –  marlow Dec 19 '11 at 1:13