Nullspace of second fundamental form when we isometrically embed $M^n$ into $R^{n+2}$ First we suppose there exists an isometric embedding $M^n \rightarrow R^{n+2}$, here $M^n$ is a n-dim Riemannian manifold. If Riemannian metric of $M^n$ is flat at a point $p$, i.e. locally there exists a coordinate charts of neighbourhood of $p$, s.t. Levi-Civita connection is locally flat $\nabla_{\frac{\partial}{\partial x_i}} \frac{\partial}{\partial x_j} = 0$.
My question is that:
Can we prove that second fundamental form of $M^n$ at point $p$ has at least $(n-2)$-dim nullspace?
Background:
As far as I tried, we can use Codazzi equation on Do Carmo's book, $K(x,y) - \bar{K}(x,y) = \langle B(x,x), B(y,y)\rangle - |B(x, y)|^2$ to prove that when existing an isometric embedding of codim 1, $M^n \rightarrow R^{n+1}$, second fundamental form is degenerate and has nullspace of dimension $(n - 1)$ by computing rank of 2-by-2 submatrices. (here $x, y$ are tangent vectors at $p$, $B(-, -)$ is second fundamental form, $K(-, -)$ is sectional curvature.) But how to prove that in case of codim 2 isometric embedding, it is degenerate and has at least $(n-2)$-dim nullspace?
 A: This question is a straightforward corollary of Proposition 1 by John Douglas Moore in his paper "Conformally Flat Submanifolds of Euclidean Space*" in 1977. I first give a short answer to it and would update it with more details soon.
In Kobayashi's book of "Foundations of differential geometry", we can prove a formula
$$R(x, y, z, w) = \langle B(x, z), B(y, w)\rangle - \langle B(x, w), B(y, z)\rangle.$$
Here $B(-, -)$ is second fundamental from $V \rightarrow W$. So $$\langle B(x, z), B(y, w)\rangle  - \langle B(x, w), B(y, z)\rangle  = 0$$ for a flat point $p$.
Let $V = T_pM$. Let $q = \max\{\dim B(z)(V): z \in V\}$. An element $x$ is regular if $\dim B(x)(V) = q$. Here $B(x)(V) = B(x, v)$ for all $v \in V$.
If $B(n, x) = 0$, from above, we get that $$\langle B(x, z), B(y, n)\rangle  = \langle B(n, x), B(y, z)\rangle  = 0.$$
Also, if $x$ is a regular element and $B(n, x) = 0$, then $tB(n, y) = B(n, ty) = B(n, x + ty) \in B(x + ty)(V)$.
When $t$ is small, $dimB(x+ty)(V) = dimB(x)(V)$, since $t \rightarrow dimB(x+ty)(V)$ is lower semi-continuous. So we can prove that $B(n, y) \in B(x)(V)$ for all $y \in V$.
Let $x$ be a regular element, $N(B, x)$ be nullspace of map $B(x)$, and $N(B)$ be the nullspace of second fundamental form $B$. We can use above to prove that $\dim N(B) \geq \dim N(B, x) \geq \dim V - \dim W$.
The last inequality comes from that $B(x): V \rightarrow W$ is a linear map and has kernel at least $\dim V - \dim W$. The first inequality follows from above by noticing that we can get $<B(n, y), v> = 0$ for all $v \in B(x)(V)$ and $B(n, y) \in B(x)(V)$ so $B(n, y) = 0 \in W$ for all $n \in N(B, x), y \in V$ .
