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Riemann curvature depends on position, on metric, and on a plane (replacing the two linear independent vector used in calculating it with any not degenerate linear combination of themselves we got the same curvature). If metric were a set of surfaces not intersecting and filling the space, the geometrical meaning would be intuitive (do a section of the surface with the plane defined by the two vector, and measure the curvature at the point of the curved line so obtained). But metric can be seen as a set of three set of surfaces intersecting at each point and filling the space, so there would be some other geometrical meaning (here for simplicity I'm supposing to be in $\mathbb{R}^3$ because I suppose this is the best way to help intuition about what Riemann curvature is, I can't visualize the general Riemann space with negative length, imaginary angles, and other things like that which I see come out in trying to generalize things).

Fausto Vezzaro
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  • The sentence "But metric can be seen as a set of three set of surfaces intersecting at each point and filling the space" makes no sense (to me). Regardless, sectional curvature can be understood using as the Gaussian curvature of surfaces which are images of planes under the exponential map. See for instance, volume 2 of Spivak's book on Differential Geometry. – Moishe Kohan Apr 24 '20 at 19:02
  • @MoisheKohan I can give the metric by writing $g_{ij}$, by writing $ds^2$ but also writing an admissible coordinate transformation, that has coordinate surfaces, I wrong? – Fausto Vezzaro Apr 25 '20 at 11:10
  • Yes, it is wrong: A Riemannian metric is fundamentally different from a coordinate transformation. – Moishe Kohan Apr 25 '20 at 13:48
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    Does this post answer your question? – C.F.G Apr 25 '20 at 14:07
  • thanks it seems useful – Fausto Vezzaro Apr 26 '20 at 08:48

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