Sectional curvature

Idea

In a surface you can compute Gaussian curvature by means of holonomy per unit area, being holonomy the rotation of a parallel transported vector which, necessarily, lies within the tangent plane of the loop (the loop is infinitesimally small, so we can look at it like contained in a plane, I guess).
In an $n$-manifold, however, even if we insist that the initial vector $w_0$ lie within the plane $\mathrm{\Pi }(u,v)$ generated by $u,v$ in order to compute the Riemann curvature tensor $R(u,v)(w)$, the parallel transported vector $w_{\parallel }(o)$ along a loop contained in $\mathrm{\Pi }(u,v)$ could stick out of $\mathrm{\Pi }(u,v)$.

But if we focus in the projection $P(w_{\parallel })$ of $w_{\parallel }$ over $\mathrm{\Pi }(u,v)$ then the rotation per unit area of $P(w_{\parallel })$ is independent of the choice of $w_0$. This quantity is called the sectional curvature of $\mathrm{\Pi }(u,v)$.

The sectional curvature determines the curvature tensor completely (seen in Wikipedia, without proof).

Definition

The sectional curvature $K(\sigma )$ of a 2-dimensional plane $\sigma$ spanned by two vectors $X$ and $Y$ in the tangent space of a manifold $M$ is given by:

where $R$ is the
If $X$ and $Y$ are orthonormal, this simplifies to:

Expression in Terms of Riemann Tensor Components

Given an orthonormal frame $\{e_1,e_2,e_3\}$, let $X=e_a$ and $Y=e_b$. The sectional curvature for the plane spanned by $X$ and $Y$ is:

Using the notation of the components of the Riemann curvature tensor, this becomes (no Einstein convention):