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I found the following equation on Wikipedia here: \begin{equation} K(u,v)={\langle R(u,v)v,u\rangle\over \langle u,u\rangle\langle v,v\rangle-\langle u,v\rangle^2} \end{equation} No explanation I could understand was given for where this formula comes from or why it represents sectional curvature.

Can someone just briefly summarize how this formula works in words? How does it let me measure sectional curvature?

Stan Shunpike
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  • Do you have some other definition of sectional curvature in mind? Otherwise, this is a definition. In absence of some other notion of what sectional curvature is or should be, this is the sectional curvature. – Muphrid Jan 30 '15 at 22:37
  • No. I was just trying to get some intuition for exactly why these particular operations are used to define sectional curvature. When you're not terrifically talented at mathematics, sometimes you have to ask a lot of questions to get understanding. – Stan Shunpike Jan 30 '15 at 22:39
  • I found this very helpful: http://math.stackexchange.com/questions/70210/is-there-any-easy-way-to-understand-the-definition-of-gaussian-curvature – Stan Shunpike Feb 05 '15 at 00:50
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    Since nobody has mentioned this yet: the denominator in the OP's formula is just the square of the area of the parallelogram spanned by $u$ and $v$ as one can see from precalculus: $\langle u,v \rangle = |u||v|\cos(\beta)$ where $\beta$ is the angle between the two vectors, so $|u|^2|v|^2 - \langle u,v \rangle^2 = |u|^2|v|^2\sin^2(\beta) = |u\times v|^2$ where $\times$ indicate the cross product. – user676464327 Apr 30 '20 at 22:33

2 Answers2

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The idea of sectional curvature is to assign curvatures to planes - the sectional curvature of a plane $\Pi$ in the tangent space is proportional to the Gaussian curvature of the surface swept out by geodesics with starting directions in $\Pi$. Intuitively you're taking a two-dimensional slice through a given plane, and measuring the classical two-dimensional curvature of this slice.

For the purpose of computation, the easiest way to represent a plane is by two vectors $u,v$ that span it. In the case where $u$ and $v$ are orthonormal, the sectional curvature is simply $\langle R(u,v)v,u\rangle$. You can determine this is the correct expression in the 2-dimensional case by showing it's equal to the Gaussian curvature, and this carries over to general dimension using the Gauss-Codazzi relations and the fact that the second fundamental form of the slice is zero at the base point of $\Pi$.

The formula you've given is in terms of an arbitrary basis for $\Pi$. It's clear that the formula must change in this case, since e.g. scaling one of the vectors linearly would change the curvature expression quadratically. To find the correct formula in an arbitrary basis, you can e.g. apply Gram-Schmidt to $u,v$ to get an orthonormal basis $e_1, e_2$, then write out $K = \langle R(e_1,e_2)e_2,e_1 \rangle$ in terms of $u,v$ and use the symmetries/bilinearity of $R$ to simplify.

Anthony Carapetis
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  • Correction: In general, there is not a totally geodesic submanifold with tangent plane $\Pi$. What you want is the $2$-dimensional submanifold $S$ swept out by geodesics starting at the point $p$ such that $\Pi\subseteq T_p M$, and with initial velocity in $\Pi$. Geodesics starting elsewhere in $S$ and tangent to $S$ might not stay in $S$. – Jack Lee Jan 29 '15 at 09:46
  • @JackLee: thanks - apparently totally geodesic surfaces are in much shorter supply than I had thought. I'll correct my answer. How strong a condition is the universal existence of totally geodesic surfaces - does it restrict us to just space forms or similar? I clearly have a hole in my knowledge here. – Anthony Carapetis Jan 29 '15 at 11:18
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    Apparently it happens only for space forms. See the answer to this MO question. – Jack Lee Jan 29 '15 at 15:46
  • Is this formula relevant to understanding the Ricci curvature tensor? According to Wikipedia: "Indeed, if $\xi$ is a vector of unit length on a Riemannian $n$-manifold, then $Ric(\xi,\xi)$ is precisely $(n−1)$ times the average value of the sectional curvature, taken over all the 2-planes containing $\xi$. There is an $(n−2)$-dimensional family of such 2-planes" and so only in dimensions 2 and 3 does the Ricci tensor determine the full curvature tensor." I am assuming $K(u,v)$ is not the average sectional curvature. Can it be used to find the average? – Stan Shunpike Jan 29 '15 at 21:29
  • To be clear, I am trying to better understand the Ricci tensor. And from what I understand, that means understanding sectional curvature. I know Ricci is the contracted version of the Riemann tensor but this didn't really give me much info about how it actually measures the curvature. This formula looked relevant so I inquired about it. But if this isn't relevant to understanding Ricci, perhaps someone more knowledgeable could point me towards more relevant information regarding how Ricci measures curvature. – Stan Shunpike Jan 29 '15 at 21:36
  • @StanShunpike: The Ricci curvature is quite literally (proportional to) the average of the $K(u,v)$: for $|\xi| = 1$, $\text{Rc}(\xi, \xi)$ is the average value of $K(\xi,v)$ when $v$ is sampled uniformly from vectors orthonormal to $\xi$. If you're looking for intuition about the Ricci curvature then this is (in my opinion) not the easiest way to think about it - I prefer to think in terms of directional volume growth, though this is harder to connect with the formula. Perhaps see http://math.stackexchange.com/questions/468651/geometrical-interpretation-of-ricci-curvature/469005#469005 – Anthony Carapetis Jan 29 '15 at 23:18
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Perhaps it's best just to look at what's going on here geometrically.

I said in one of your other questions that the Riemmann tensor and sectional curvature can be written directly in terms of the exterior algebra. Namely, given a simple 2-vector $B$,

$$K(B) = \frac{\langle R(B), B \rangle}{\langle B, B \rangle}$$

where the inner product $\langle,\rangle$ is extended as usual to the space of 2-vectors.

With this in mind, you can think of $R$'s matrix representation on the space of 2-vectors. We're taking one of the elements on the diagonal--$\langle R(B), B \rangle$--and normalizing based on the magnitude of $B$--that is, by $\langle B, B \rangle$--so that the function $K$ can be thought of as a function of planes, independent of the magnitude of the 2-vector $B$ plugged into it.

The key in comparing this to the regular formula is to recognize the denominator as the inner product of 2-vectors (of two planes with each other) and the numerator similarly.

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Muphrid
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