# Sectional curvature in a paraboloid is always positive.

I'm working on Lee's book ''Riemaniann Manifolds an Introduction to Curvature''. One exercise (11.1) is about to see that the paraboloid given by the equation $y=x_1^2+...+x_n^2$ has positive sectional curvature everywhere.

It is known that $K(\pi)=\frac{\langle R(u,v)v,u \rangle}{\langle u,u \rangle \langle v,v \rangle - \langle u,v \rangle ^2}$ where $\pi$ is the plane generated by $u,v$. Cauchy Schwarz ensures denominator is always positive so it would be enough to check that $\langle R(u,v)v,u \rangle$ is positive at any point for any pair of tangents vectors. That is equivalent to check $\langle\nabla_X \nabla_Y Y - \nabla_Y \nabla_X Y -\nabla_{[X,Y]}Y,X\rangle$ is positive where $\nabla$ is Levi-Civita connection. The first Christoffel identity (here) https://en.wikipedia.org/wiki/Fundamental_theorem_of_Riemannian_geometry allows to compute the Levi-Civita connection via Christoffel symbols. I can compute the Christoffel symbols by finding the metric of the paraboloid given by the evident chart (just by pull-back).

The question is if there is an easier way to check the paraboloid has positive sectional curvature everywhere or you have to follow the path I talked about; that I do not find complicated, but it requires some work with many calculations.

• You can pick $u$ and $v$ to be orthogonal unit vectors and therefore get rid of the denominator in expression for $K$ Commented Jan 3, 2016 at 10:45
• Denominator was not a problem since Cauchy Schwarz ensured that denominator was always positive. Now I'm thinking about using en.wikipedia.org/wiki/Second_fundamental_form with the Gauss Equation. However, I am still interested in other ways to solve the problem.
– DCao
Commented Jan 3, 2016 at 18:13
• Using the first and second fundamental forms is an option, but from the point of computational work it is not much easier Commented Jan 3, 2016 at 21:14

The key is using that the paraboloid is a submanifold of $\mathbb{R}^n$ combined with Gauss Equation.
Because we are working on an Euclidean submanifold (as we posted in the comments) if $u=\alpha_1 e_1+...+\alpha_{n+1} e_{n+1}$, $v=\beta_1 e_1+...+\beta_{n+1} e_{n+1}$ and $N$ is a normal unitary vector field (for example, $(2x_1,...,2x_n,-1)/f$ (where $f$ is the norm of the numerator) the shape operator gives $S(u)=\nabla_u N=(2 \alpha_1,..., 2\alpha_n,0)/f + u(1/f)fN$ where $\nabla$ is Levi-Civita in $\mathbb{R}^n$(so Christoffel Symbols are $0$). Now $K(\pi)$ has the same sign that (we also use $\langle N,u \rangle = \langle N,v\rangle=0) \rangle$$\langle S(u),u \rangle \langle S(v),v \rangle - \langle S(u),v \rangle^2 =4(\alpha_1^2+...+\alpha_n^2)(\beta_1^2+...+\beta_n^2)-4(\alpha_1 \beta_1+...+\alpha_n \beta_n)^2 \geq 0$$ Finally we should show that equality is impossible. If the equality holds then$u$and$v$are proportional in the first$n$coordinates. The condition$\langle u, N\rangle=\langle v, N\rangle=0$makes that, in that case, the last coordinate has the same proportion too (because the last coordinate of N never vanishes). But in that case$u$and$v\$ are proportional and can not generate a plane.