# Sectional curvature of hypersurface

Let $M$ be a Riemannian hypersurface in $\mathbb{R}^{n+1}$, where $n \geq 2$. Let $\varSigma$ be a two dimensional subspace of the tangent space $T_{p}M$ of $M$ at $p$, and denote by $s_{p}$ the shape operator of $M$ at $p$.

I believe I read on some reference (which currently I am not able to retrieve) the following statement:

The sectional curvature $K(\varSigma)$ is equal to the determinant of the restriction of $s$ to $\varSigma$, provided $\varSigma$ is $s$-invariant, i.e. $s(\varSigma) \subset \varSigma$.

Can somebody confirm that such statement is true? How would you prove it?

Yes, this is correct. Start with an orthonormal moving frame $e_1,\dots,e_n; e_{n+1}$ with $e_{n+1}$ the normal, and let $\omega_i$ be the dual coframe on $M$. If $e_1,e_2$ is an orthonormal basis for $\Sigma$ and $\Omega_{ij}$ is the curvature $2$-form, we have $$K(\Sigma) = -\Omega_{12}(e_1,e_2).$$ By the Gauss equation, we also have $$\Omega_{12} = -\sum_{i,j=1}^n h_{1i} h_{2j} \omega_i\wedge\omega_j,$$ where $(h_{ij})$ is the matrix of the second fundamental form. In particular, we see that $K(\Sigma) = h_{11}h_{22}-h_{12}h_{21}$, and this is precisely the determinant of $s|_\Sigma$ in the event that $s(\Sigma)\subset\Sigma$. (After all, the matrix of $s_p$ with respect to the basis $e_1,\dots,e_n$ is precisely $(h_{ij})$.)
By the way, here's something else that you may not know relating to the sectional curvature. $K(\Sigma)$ is precisely the Gaussian curvature at $p$ of the surface $\exp_p(\Sigma)$.