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In his book Riemannian Geometry, Manfredo Do Carmo states the following on page 97:

Let $x = z_n$ be a unit vector in $T_pM$; we take an orthonormal basis $\lbrace z_1,z_2,...,z_{n-1}\rbrace$ of the hyperplane in $T_pM$ orthogonal to $x$ and consider the following averages: \begin{equation} \text{Ric}_p(x) = \frac{1}{n-1} \sum_i \langle R(x,z_i)x,z_i \rangle \end{equation} for $i = 1,2,...,n-1$

I know $R(\phantom{x},\phantom{x})$ is the Riemann curvature tensor. Presumably then, $R(x,z_i)x$ measures the noncommutativity of parallel transporting a vector field $x$ in the $z_i$ direction and then the $x$ direction. Something like \begin{equation} R(x,z_i)x = \nabla_{z_i} \nabla_{x}x- \nabla_{x} \nabla_{z_i}x \end{equation}

My question:

What does the inner product $ \langle R(x,z_i)x,z_i \rangle$ represent geometrically?

Stan Shunpike
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    In an orthonormal basis, $\left \langle R\left(x, z_{i}\right)x, z_{i}\right \rangle$ represents the component of $R(x, z_{i})x$ in the direction of $z_{i}$ (i.e., at a point, the scalar projection of the vector $R(x, z_{i})x$ onto $z_{i}$). – THW Feb 02 '15 at 15:32

1 Answers1

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I think it's easiest to first give a description of what $R(X,Y)Z$ represents geometrically. To be precise, let $p \in M$ and let $u,v,w \in T_pM$; we'll give a geometric description of $R(u,v)w$. To do this, extend $u$ and $v$ to coordinate fields $U, V$ on a neighborhood of $p$. Thus I'm supposing we have a chart $\phi : U \to M$, where $U \subseteq \mathbb{R}^n$ such that $\phi(0) = p$ and $d\phi(\partial_1)_p = u, d\phi(\partial_2)_p = v$.

Then let $\sigma$ denote the square of sidelength $\varepsilon$ in $\mathbb{R}^n$ tracing out the verticies $$ (0,0) \to (\varepsilon, 0) \to (\varepsilon, \varepsilon) \to (0, \varepsilon) \to (0,0) $$ in that order, and let $\gamma = \phi \circ \sigma$ be the image of this square in $M$. Then if $P_\gamma$ denotes parallel transport around $\gamma$, one can show $$ P_\gamma w = w - \varepsilon^2 R(u, v)w + O(\varepsilon^3) $$ where I don't guarantee the sign is correct. So $R(u,v)w$ gives the first-order correction to parallel translation around small coordinate loops, and the correction is proportional to the "area" of the loop.

Now, in the case that $M$ is two-dimensional and let $u, v$ be orthonormal vectors at $p$. Then by our formula above $$ \langle P_{\gamma} u, v \rangle = -\varepsilon^2 \langle R(u,v)u, v \rangle + O(\varepsilon^3). $$ On the other hand, $\langle P_\gamma u, v \rangle$ is equal to $\sin \theta_\gamma$, where $\theta_\gamma$ is the angle change due to parallel translate around our path $\gamma$ -- note that in this two-dimensional case $\theta_\gamma$ completely describes the path-dependence of parallel transport. By definition $\langle R(u,v)u, v \rangle$ is the sectional curvature $\kappa$ of $M$. So our formula says roughly that $$ \theta_\gamma \approx - A(\gamma) \kappa, $$ where $A(\gamma)$ is the area inside $\gamma$ and the approximation holds for small curves $\gamma$, and I don't guarantee the sign (which depends on defining the sign of $\theta_\gamma$ precisely anyway).

One thing I like about this picture is that it makes it more or less obvious that something like the Gauss-Bonnet formula should hold, since to parallel transport around a big loop in a surface $M$ you can instead imagine breaking its inside into small loops and parallel transporting around those.

In fact the Gauss-Bonnet formula is a good way of making precise for surfaces this intuition that the sectional curvature $\kappa(u,v)$ of the plane spanned by $u$ and $v$ measures the path dependence of parallel transport for small curves living in an $\mathbb{R}^2$ coordinate chart through $u$ and $v$.

You can also instead think of the sectional curvature $\kappa(u,v)$ (which is just $\langle R(u,v)u, v \rangle$) as a Gaussian curvature, which is a picture that better explains the use of the term "curvature". (In general $\kappa(u,v)$ is the Gaussian curvature of the surface swept out by geodesics whose initial tangent vectors lie in the plane spanned by $u$ and $v$.) The "informal picture" of Gaussian curvature given on Wikipedia is a good picture for this viewpoint.

mollyerin
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  • +1 thanks for replying. Okay, but I thought sectional curvature had to be $$K(u,v) = \frac{\langle R(u,v)u,v \rangle}{|u \wedge v|^2}$$? What about the square of the area of the parallelogram I am supposed to divide by? Or is that incorporated because $u$ and $v$ are orthonormal? – Stan Shunpike Feb 02 '15 at 20:11
  • As you suggest, the denominator is $1$ when $u$ and $v$ are orthonormal. Your expression is nice though, as it gives the sectional curvature for the plane spanned by $u$ and $v$ regardless of what $u$ and $v$ are (provided they're linearly independent, of course). Of course, in the formula you have for the Ricci tensor the vectors are orthonormal, so thinking of the summand as a sectional curvature is definitely a good idea. – mollyerin Feb 02 '15 at 20:53
  • Does Ricci curvature always involve orthonormal vectors? Because if it does, then the denominator always equals 1. I didn't realize that was part of defining Ricci curvature. I assumed $u$ and $v$ need not be orthonormal. Obviously, there is no such requirement for sectional curvature in general. So why would there be such a requirement when measuring Ricci curvature? – Stan Shunpike Feb 02 '15 at 23:05
  • No, the Ricci tensor is, in particular, a tensor, so you can apply it to any two vectors you want, and it's bilinear in its arguments. (Note that $K(u, v)$ is not a tensor; it doesn't scale linearly with its arguments.) But if you want to interpret $\text{Ric}_p(u, u)$ as an average of sectional curvatures, $u$ had better be a unit vector, so that the terms appearing in your formula look like sectional curvatures. – mollyerin Feb 03 '15 at 09:00