Sectional curvature

This item assumes ${\mathrm{A}\mathrm{C}}_{\omega }$, namely countable choice. In the propagated dependency chain, that assumption is required through Algebraic symmetries of the Riemann tensor; after those interfaces are fixed, the remaining local or finite argument makes no additional countable-family choice.

Let $\sigma \subseteq T_{p}M$ be a two-dimensional linear subspace and let $(X,Y)$ be any ordered basis of $\sigma$. Its sectional curvature is

$K(\sigma ):=\frac{\mathrm{Rm}(X,Y,Y,X)}{g(X,X)g(Y,Y)-g(X,Y{)}^2}\mathrm{.}$

The denominator is the Gram determinant of the independent pair $(X,Y)$ and is strictly positive because $g$ is positive definite. The next lemma proves that the quotient is unchanged by the chosen ordered basis. With the sign convention fixed above, an orthonormal tangent two-plane in the unit round sphere has sectional curvature $+1$.

There are no tangent two-planes in dimensions zero or one, so the definition has empty domain there rather than assigning a spurious value. No plane exists on an empty manifold either. The same fibrewise definition applies at a boundary point.