It is well-known that any point $p$ of a surface $M^{2}$ in $\mathbb{R}^{3}$ is either elliptic, hyperbolic, parabolic, or planar. These notions are defined in terms of the principal curvatures, i.e., the eigenvalues of the shape operator.
In particular, for any $p\in M^{2}$, the second fundamental form $\alpha$ is either zero or has at most two asymptotic directions. One observes that the set of asymptotic vectors of $\alpha$ is the union of vector subspaces of $T_{p}M^{2}$. (A vector $v\in T_{p}M^{2}$ is asymptotic if $\alpha(v,v)=0$.)
I am wondering what happens if we increase the dimension and consider an $m$-dimensional hypersurface $M^{m}$ of $\mathbb{R}^{m+1}$: Can the asymptotic set still be decomposed as a union of subspaces of $T_{p}M^{m}$?
So far I tried to derive this result from the generalized Euler's formula: If $\kappa_{1},\dotsc,\kappa_{m}$ are the principal curvatures corresponding to the principal directions $v_{1},\dotsc,v_{m}$, then the normal curvature $\alpha(v,v)$ is given by:
\begin{equation}
\alpha(v,v)=\sum_{i=1}^{m}\kappa_{i}\cos(\theta_{i})^2\,,
\end{equation}
where $\theta_{i}$ denotes the angle between $v$ and $v_{i}$.
Best Answer
The answer to your question is no. In general, the set of asymptotic vectors will be a quadratic cone (and the dimension of the linear rulings of this cone will depend on the rank of the second fundamental form). If the rank of the second fundamental form is $2$ (and the form is indefinite), then you will get a union of hyperplanes, but if the rank is higher, the you will have a "curved cone" ruled by linear spaces of dimension depending on the rank.
By the Spectral Theorem, you can find an orthonormal basis for the tangent space so that, in the new coordinates, we have $\text{II}(x,x) = \sum\lambda_i x_i^2$, and you can deduce everything from knowing how many positive $\lambda_i$, how many negative $\lambda_i$, and how many zero $\lambda_i$ you have.