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The sectional curvature is defined as

$$K = \frac{g(R(x,y)z,t)}{g(x,x)g(y,y) - g(x,y)^2 }$$

It is said (and proven in Da Carmo) and well known that the sectional curvature can completely determine Rieman curvature tensor $R$ for any hyperplane spanned by $(x,y)$.

I am not sure what the purpose of this sectional curvature is, but to determine $K$, don't you need to know $R$ in the first place? And then you can compute the numerator with $g$. I don't understand the point taking the (inner product) of $R$ with say $t$ ($R(X,Y,Z,T)$). What is the purpose of this other than to exploit some symmetries (e.g. Bianchi identities)

Lemon
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  • See if my answer here, the question, the comments, and the links therein can be helpful. – Yuri Vyatkin Sep 16 '18 at 03:57
  • For the geometric meaning (interpretation) of various curvature quantities, see this question. – Yuri Vyatkin Sep 16 '18 at 07:13
  • Another very relevant question with a good answer is here. – Yuri Vyatkin Sep 16 '18 at 07:18
  • Moreover, your question sounds like a duplicate of this one (which has a good answer too). – Yuri Vyatkin Sep 16 '18 at 07:21
  • @YuriVyatkin i've read them. But all I ever understood from them is that the Ricci is a "simplification" (contraction) of the Riemann Curvature Tensor. Basically by sacrificing information (averaging basically), we get symmetry. The Sectional Curvature coincides with Gaussian Curvature and supposedly a "generalization" of it. Scalar curvature is a further simplification (second contraction or trace) of Ricci (2nd averaging). I guess my real question was, just why? I actually find the Rieman tensor way more intuitive than the other curvatures. – Lemon Sep 26 '18 at 09:20

1 Answers1

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As stated, the original question is not clear enough to attract attention. From the discussion in the comments I understand that the real question of the OP must be

Why the sectional curvature is important (interesting)?

I don't have much time for writing this up (and I need to brush up my memory on this topic), but I will take a risk to leave a few lines as a sketch for a possible more complete answer:

  • the Riemann tensor has values in a more complicated space, whereas the sectional curvature has numeric values, that makes it more convenient for various comparisons;
  • values (signs and bounds) of the sectional curvature affect or determine the topology of the manifold (see, e.g. John Lee, Riemannian Geometry: Introduction to Curvature, Chapter 11 "Curvature and Topology);
  • the sectional curvature can be computed without prior knowledge of the Riemann tensor, e.g. from the coefficients in Taylor expansion of some geometric quantities.

To support the last claim, here are a couple of references for further reading:

[1] Yann Ollivier, A visual introduction to Riemannian curvatures and some discrete generalizations, http://www.yann-ollivier.org/rech/publs/visualcurvature.pdf, p. 3 (the sectional curvature arises as a coefficient in the Taylor expansion of the geodesic deviation).

[2] Ryan Budney, Differential Geometry lecture notes, 2017, http://rybu.org/node/79, p. 108 (the sectional curvature arises as a coefficient in the Taylor expansion the length of the geodesic circle).

Perhaps, there are more authoritative sources and precise statements somewhere else, and if I find them I may update this answer (or, someone else can do this too).

Yuri Vyatkin
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  • Thank you for your effort. I actually hit "enter' before I could finish my title. I didn't realize my mistake until much later. – Lemon Sep 26 '18 at 22:03