For some hypersurface $M \subset \mathbb{R}^n$, with $n \geq 4$, I want to show that for the sectional curvature of $M$, it holds that it can't be negative.
I somehow don't know how to start here. I mean I know that:
$K(span(v,w))=\overline{K}(span(v,w))+ \Vert \nu \Vert \dfrac{<Sv,v><Sw,w>-<Sv,w>^2}{<v,v><w,w>-<v,w>^2} $
where $\overline{K}(span(v,w))$ is the sectional curvature of $\mathbb{R}^N$ and $\nu$ is the unit normal on $M$. Also, I know that $\overline{K}(span(v,w))=0$ and that $<v,v><w,w>-<v,w>^2>0$
Now $Sv=\overline{\nabla}_v \nu$ right? I just don't know how to use the properties of $\mathbb{R}^n$ here. How does the Levi-Civita connection look like for $\mathbb{R}^n$?
As a hint, I have that its possible to diagonalize $S$, but don't really understand this.
Here is also a picture of the exercise:
The diagonalization hint is a nice way to go. Since $S: T_pM\to T_pM$ is symmetric, there is a orthonormal basis for $T_pM$ with $Se_i =\lambda_i e_i$ and $\lambda_i\in\mathbb{R}$.
Now for $i\ne j$ you have $K(e_i, e_j) = \lambda_i\lambda_j$.
If some $\lambda_i$ is zero you are done (then $K(e_i, e_j) = 0$ for this $i$).
So assume you have at least three $\lambda_1, \lambda_2, \lambda_3$ all non-zero. Then at least two of these have the same sign (say $\lambda_1, \lambda_2$ have the same sign) and so the sectional curvature through this plane ($e_1, e_2$) will be positive.