3
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 $z_{i}$). $\endgroup$ – THW Feb 2 '15 at 15:32
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ +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? $\endgroup$ – Stan Shunpike Feb 2 '15 at 20:11
  • $\begingroup$ 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. $\endgroup$ – mollyerin Feb 2 '15 at 20:53
  • $\begingroup$ 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? $\endgroup$ – Stan Shunpike Feb 2 '15 at 23:05
  • $\begingroup$ 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. $\endgroup$ – mollyerin Feb 3 '15 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.