# Interpretation of Ricci and Scalar curvature.

Let $$(M,g)$$ be a SR-manifold and for $$p \in M$$, let $$\{e_1,\ldots,e_n\}$$ be a pseudo-orthonormal basis for $$T_pM$$. Now, in my notes, it says that, if $$v \in T_pM$$ is such that $$g(v,v) \neq 0$$ and $$\{e_2,\ldots,e_n\}$$ is an orthonormal basis of $$v^{\perp}$$ then $$\operatorname{Ric}(v,v) = g(v,v) \sum_{i = 2}^{n} K(v,e_i)$$ where $$K(-,-)$$ is the sectional curvature of non-degenerate two-planes at $$p \in M$$. Then it says that from this, we can interpret $$\operatorname{Ric}(v,v)$$ as the ”mean” of all sectional curvatures of two-planes containing $$v$$.

Now, for the scalar curvature $$\operatorname{scal} = \operatorname{tr}_g(\operatorname{Ric}) \in C^{\infty}(M)$$ we have $$\operatorname{scal}(p) = \sum_{i,j = 1}^{n} \epsilon_i \epsilon_j R(e_j,e_i,e_i,e_j) = \sum_{\substack{i,j = 1, \\ i \neq j}}^{n} \epsilon_i \epsilon_j K(e_i,e_j)(\underbrace{g(e_i,e_i)}_{=\epsilon_i}\underbrace{g(e_j,e_j)}_{=\epsilon_j}-\underbrace{g(e_i,e_j)}_{= 0}) = \sum_{i \neq j} K(e_i,e_j)$$ so that $$\operatorname{scal}(p)$$ is the ”mean” over all $$i \neq j$$ sectional curvatures of non-degenerate two-planes.

My question is this: In what sense are these two descriptions ”means”? My mind is naturally drawn to the arithmetic mean, but there are of course other means, and I am not sure there is a mathematical definition of mean independent of the specific mean one is referring to, but since we don’t divide by something, it can’t be the arithmetic mean, I presume. Any thoughts on in which sense these are means? And for the scalar curvature, I presume they must mean all non-degenerate two-planes spanned by the basis vectors? Because could there not potentially be other two-planes?

It's the arithmetic mean, up to a constant multiple. If $$v$$ is a unit vector, then $$\operatorname{Ric}(v,v)/(n-1)$$ is the average of the sectional curvatures of basis $$2$$-planes containing $$v$$. Similarly, if you divide the scalar curvature by $$n(n-1)$$, you get the average of all the sectional curvatures of basis $$2$$-planes.

You can turn this into a more invariant kind of average that doesn't depend on a choice of basis: If you think of the set of all $$2$$-planes containing a unit vector $$v\in T_pM$$ as being parametrized by the set of unit vectors orthogonal to $$v$$ (an $$(n-2)$$-sphere), then you can show that $$\operatorname{Ric}(v,v)/(n-1)$$ is equal to the integral of sectional curvatures over that sphere divided by the volume of the sphere. Similarly, $$\operatorname{scal}/(n(n-1))$$ is equal to the integral of sectional curvatures over the Grassmannian of oriented $$2$$-planes in $$T_pM$$, divided by the volume of that Grassmannian.

• Hm, are there not $$\binom{n}{2} = \frac{(n)(n-1)}{2}$$ choices, so that we should divide the scalar curvature by $$\frac{(n)(n-1)}{2}?$$ Aug 20, 2023 at 21:37
• Anyway, I will accept this answer, since I think this made it easy to understand what they meant. I have heard the expression ”Grassmanian” but I am not so familiar with it. I am possibly also somewhat starstruck, since I have read several of your books, so I might be biased . Aug 20, 2023 at 21:46
• @Ben123: In the formula for the scalar curvature (as the trace of the Ricci tensor), each 2-plane shows up twice -- For example, the span of $(e_1,e_2)$ appears in the expression for $\operatorname{Ric}(e_1,e_1)$, and again in the expression for $\operatorname{Ric}(e_2,e_2)$. Aug 21, 2023 at 4:30
• I am just going from the expression $$\operatorname{scal}(p) = \sum_{i \neq j} K(e_i,e_j)$$ and disregarding order (since $K(e_i,e_j) = K(e_j,e_i)$) and for each choice we have to pick $2$ basis vectors out of $n$ basis vectors. Combinatorially that is precisely $n$ choose $2$. Aug 21, 2023 at 4:48
• Yes, but $K(e_1,e_2)$ and $K(e_2,e_1)$ both appear in that sum, so each $2$-plane is counted twice. Aug 21, 2023 at 15:18

You can apply the fact that $$tr(b)= \frac1{n|\partial B_1(0)|}\int_{\partial B_1(0)} b(v,v) dv$$ for a symmetric bilinear form $$b:V\times V\to\mathbb{R}$$, to the setting of $$V= T_xM$$. This yields multiple results, depending on what bilinear form you use. For example, one prove that $$S(x)= \frac1{|\partial B_1(0)|}\int_{\partial B_1(0)} Ric(v,v) dv$$. Similarly, $$Ric(v,v)= \frac{n-1}{|S|}\int_{S} K(v,u) du$$ with $$S=\{u\in T_xM: \Vert u \Vert = 1,\, u\perp v\}$$. The constants depending on $$n$$ depend on the definitions that you are using.

This shows, that there is an analytic meaning behind the word „mean“. It is essentially the arithmetic mean of the eigenvalues of the matrix related to symmetric bilinear form that you are studying. (There might be a factor depending of $$n$$ infront of the arithmetic mean, but that doesn’t matter that much).