Should minimal hypersurface have nonpositive scalar curvature? Let $\Sigma^n$ be an embedded minimal hypersurface of an orientable Riemannian manifold $M^{n+1}$, then the mean curvature $H$ of $\Sigma$ vanishes everywhere. Let $\lambda_1,…\lambda_n$ be those principal curvatures of $\Sigma$. Then, $$0=H^2=\sum_i\lambda_{i}^{2}+\sum_{i\neq j}\lambda_i\lambda_j.$$ We know that each $\lambda_i\lambda_j$ is exactly the sectional curvature along the space spanned by the principal directions corresponding to these two principal curvatures. Hence, the scalar curvature of $\Sigma:$ $$S=\sum_{i\neq j}\lambda_i\lambda_j\leq0.$$
This is quite ridiculous for me. In fact, if the metric of $\Sigma$ is given by $g_{ij}dx_idx_j$ and we let $M=\Sigma\times\mathbf R$ with the metric $g_M=g_{ij}dx_idx_j+dt^2.$ Then, $\Sigma\times\{0\}$ is a minimal hypersurface of $M$ since the zero function satisfies the minimal surface equation on $\Sigma$ and its graph is $\Sigma\times\{0\}$. If $\Sigma$ has positive scalar curvature somewhere, so does $\Sigma\times\{0\}$.
 A: You're correct that this is absurd; the error is in your definition of sectional curvature, discussed below. The computation about principal curvatures is correct, and gets at some interesting phenomena exhibited by minimal submanifolds. For instance, all embedded minimal $2$-manifolds of Riemannian manifolds are "saddle-shaped" in the sense that $\lambda_1 \lambda_2 \leq 0$, and as a consequence their sectional curvatures are bounded above by the sectional curvatures of the ambient space.
In the case of embedded $2$-dimensional submanifolds $(M^2, g)$ of $\mathbb{R}^3$, the scalar curvature of $M$ at $p$ is (twice) the product of the principal curvatures at $p$, which is independent of the realization of $(M^2,g)$ as a Riemannian submanifold of $\mathbb{R}^3$ by Gauss's Theorem Egregium. In this case, the observation is fully correct: every embedded minimal surface in $\mathbb{R}^3$ has nonpositive scalar curvature everywhere.
In general though, the products of principal curvatures of a Riemannian submanifold $(M,g) \subset (M',g')$ depend heavily on the realization of $(M,g)$ inside $(M',g')$, and upon $(M',g')$. Accordingly, the sectional curvatures of $M$ are defined purely in terms of $(M,g)$, and not in terms of an embedding of $(M,g)$ elsewhere.
