# Ricci tensor and average of a tensor

Let $(M^n,g)$ be an oriented Riemannian $n$- manifold and $g$ is a Riemannian metric on $M$ , $\mathrm{d}\sigma$ is Riemannian volume form on $S^{n-1}$ and $\text{Vol}(S^{n-1})$ is volume of $S^{n-1}$.

I have proved that the average of a symmetric 2-tensor $\alpha(V,V)$ on all unit vectors $V\in S^{n-1}\subset T_pM^n$ is $\dfrac{1}{n}$ of trace of $\alpha$. i.e.

$\dfrac{1}{n}\text{trace}_g(\alpha)=\dfrac{1}{\text{Vol}(S^{n-1})}\displaystyle\int_{S^{n-1}}\!\alpha(V,V)\mathrm{d}\sigma$.

From this, How can I show that for any unit vector $U$ that $\dfrac{1}{n-1}\mathrm{Ric}(U,U)$ (Ricci tensor) is the average of the sectional curvatures of planes containing the vector $U$.

Let $\mathrm{R}$ is Riemannian tensor and $\mathrm{K}$ is sectional curvature. I defined that $\alpha(Y,Z):=\mathrm{R}(U,Y,Z,U)$.It is clear that $\alpha$ is symmetric 2-tensor. Now $\alpha(V,V)=\mathrm{K}(U,V)$.How does it Work? What's your ideas?

First of all you need to make sense of "averaging along all planes containing $U$". This can be done: Let $U^\perp \subset T_pM$. Then all planes containing $U$ are in one-to-one correspondence with the unit sphere $S^{n-2} \subset U^\perp$.

Thus the average $A$ is given by

$$A = \frac{1}{Vol(S^{n-2})} \int_{S^{n-2}} \ \mathrm R(U, V, U, V) d\sigma = \frac{1}{Vol(S^{n-2})} \int_{S^{n-2}} \ \alpha(V, V) d\sigma.$$

Now from what you have proved (well, not exactly what you proved), we have

$$\begin{split} \frac{1}{n-1} Ric (U, U) &= \frac{1}{n-1} \left(R(U, U, U, U) + \sum_{i=2}^n R(U, e_i, U,e_i)\right) \\ &=\frac{1}{n-1} \sum_{i=2}^n R(U, e_i, U,e_i)\\ &= \frac{1}{n-1} tr (\alpha|_{U^\perp})\\ &= A \end{split}$$

• 1. $U^\perp$ is by definition the set of all vectors in $T_pM$ perpendicular to $U$. 2. Let $L$ be a two plane containing $U$, then there is a unit vector $V$ in $L$ so that $V$ is perpendicular to $U$ and $V$ is of norm one. This vector lie in $V \in S^{n-2} \subset U^\perp$. @bigli – user99914 Jun 9 '15 at 4:00
• @John1.If your mean from $U^{\bot}$ is all vectors that perpendicular to $U$ then $U^{\bot}\not\subset T_pM$. Since $U\in T_pM$. 2. all planes containing $U$ are in one-to-one correspondence with the unit sphere $S^{n-2}\subset U^{\bot}$, WHY? – bigli Jun 9 '15 at 4:03
• Sorry, my comment was before your first comment. Hence 1. OK. 2. Why $n-2$? – bigli Jun 9 '15 at 4:09
• That's because $U^\perp$ is $n-1$ dimensional and we consider the unit sphere in $U^\perp$, which is $n-1 -1$ dimensional. @bigli – user99914 Jun 9 '15 at 4:17
• That's true since $e_2, \cdots e_n$ are orthonormal frame in $U^\perp$ and $\alpha (V, V) = R(U, V, U, V)$. – user99914 Jun 9 '15 at 4:27