# Sectional curvature independent of basis.

Given a semi-Riemannian manifold $$(M,g)$$; for $$p \in M$$, we define the sectional curvature of a non-degenerate $$2$$-plane $$\sigma$$ with basis $$\{u,v\}$$ as $$K(\sigma) := K(u,v) = \frac{R(u,v,v,u)}{Q(u,v)}$$ where $$R$$ is the Riemann-curvature tensor and $$Q(u,v) := g(u,u)g(v,v)-g(u,v)^2.$$

We set $$\tilde{u} = au+bv$$ and $$\tilde{v} = cu+dv$$. Could someone guide me through the computation that $$Q(\tilde{u},\tilde{v}) = (ac-bd)^2Q(u,v)?$$

It's quite simple if you use the fact that the riemannian metric is, in fact, a metric. That is, its bilinearity: \begin{align} Q(\overline{u}, \overline{v}) &= Q(au+bv,cu+dv) \\ &= g(au+bv,au+bv)g(cu+dv,cu+dv) - g(au+bv,cu+dv)^2 \\ &= (a^2g(u,u) + 2abg(u,v) + b^2g(v,v)) \cdot (c^2g(u,u) + 2cdg(u,v) + d^2g(v,v)) \\ &- (acg(u,u) + (bc+ad)g(u,v) + bdg(v,v))^2 \end{align}
• The part about "by a simple calculation" could well have referred to the analogous calculation to show that $$R(\tilde{u},\tilde{v},\tilde{v},\tilde{u}) = (ad-bc)^2R(u,v,v,u).$$ Aug 7, 2023 at 14:54
$$Q(\tilde{u},\tilde{v}) = Q(au+bv,cu+dv) = \cdots = (a^2g_{u,u}+2abg_{u,v}+b^2g_{v,v}) \cdot (c^2g_{u,u}+2cdg_{u,v}+d^2g_{v,v})-(acg_{u,u}+(bc+ad)g_{u,v}+bdg_{v,v}) = \cdots = ((ad)^2-2abcdg+(bc)^2)g_{u,u}g_{v,v}-((bc+ad)^2-4abcd)g_{u,v}^2 = (ad-bc)^2g_{u,u}g_{v,v}-((ad)^2+(bc)^2+2acbd-4acbd))g_{u,v}^2 = (ad-bc)^2g_{u,u}g_{v,v}-(ad-bc)^2g_{u,v}^2 = (ad-bc)^2(g_{u,u}g_{v,v}-g_{u,v}^2) = (ad-bc)^2Q(u,v)$$