Zeros of the second fundamental form Let $ f:M \rightarrow N $ be a minimal immersion (of arbitrary codimension or an hypersurface if it is necessary) and let $ |A| $ be the norm of its second fundametal form.If $ A $ is not identically zero is it true that the zeros of $ |A| $ are isolated?
If $ f:M \rightarrow R^3 $ is a minimal immersion then the conjecture above is true. Briefly this case follows from the fact that $ f $ can be locally represented as a conformal minimal immersion $ X: \Omega \subset R^2 \rightarrow R^3 $ and for conformal minimal immersions the statement above is true (see Osserman 'Minimal surfaces')
Thanks
 A: My conjecture is that this may not be true, but in order to justify it one needs to get down to some calculations that I don't have much time to carry on. Neither I could find any relevant known fact related to that.
To make some progress in the understanding of the question I would like to post the following incomplete considerations, and see if they initiate a discussion that may bring us closer to the solution.
Let us consider a simpler case when $\mathrm{N} = \Bbb R^3$, and $\mathrm{M} \subseteq \Bbb R^2$, so we have a classical situation with a hypersurface $\mathrm{M}$ in the 3-dimensional Euclidean space.
The Bonnet theorem guarantees that if $g_{i j}$ and $\mathrm{II}_{i j}$ are such that the Gauss and Codazzi equations are formally satisfied:
$$
\mathrm{R} = \mathrm{II} \wedge \mathrm{II} \tag{Gauss}
$$
$$
\nabla \wedge \mathrm{II} = 0 \tag{Codazzi}
$$
then there exist an immersion $f \colon \mathrm{M} \to \Bbb R^3$ such that $g$ is the first fundamental form of $f$, and $\mathrm{II}$ is the corresponding second fundamental form. Of course, we have to require that $g_{i j}$ is positive definite.
(Don't worry too much about the notation in the equations above. It is enough to know that treating $g$ formally as a metric tensor we can construct the corresponding Levi-Civita connection $\nabla = \nabla^g$ on $\mathrm{M}$ which (formally) has the Riemannian curvature $\mathrm{R} = \mathrm{R}^g$. Using the Euclidean metric in $\Bbb R^3$ we can raise or lower indices freely, so $\mathrm{II}$ can be thought as a vector-valued 1-form, and we can use the exterior product and the exterior covariant derivative as shown)
In components, the first and the second f.f., or their prototypes, can be written in the classical notation as matrices
$$
g = 
\begin{pmatrix}
E & G \\
G & F
\end{pmatrix}
$$
and
$$
\mathrm{II} = 
\begin{pmatrix}
L & M \\
M & N
\end{pmatrix}
$$
where $E$, $G$, $F$, $L$, $M$, $N$ are some functions on manifold $\mathrm{M}$.
Now, let $A^{i}{}_{j} = g^{i k} \mathrm{II}_{i k}$ be the shape operator, so that $|A|^2 = A^{i}{}_{j} A^{j}{}_{i}$.
Restricting to the case of minimal manifolds ($H= g^{i j} \mathrm{II}_{i j} = 0$) gives some more polynomial conditions on variables $E$, $G$, $F$, $L$, $M$, $N$.
We also could use more linear algebra (simultaneous diagonalization of two matrices), and so on...
In general, without requiring some sort of analiticity, there is no obvious reason that we cannot make all components of $\mathrm{II}$ vanishing in a neighborhood of a point.
[To be continued if there is any interest, or new ideas pop up...]
