Let $U^{m} \subset \mathbb{R}^{m}$ be an open set. Suppose $\varphi$ is an immersion of $U^{m}$ into $\mathbb{R}^{m+n}$ satisfying the following condition:


For each point $p \in \varphi(U^{m})$, the nullity of the second fundamental form of $\varphi(U^{m})$ is equal to $m-1$. 


Then, it is well-known that


1. $\varphi(U^{m})$ is foliated by $(m-1)$-planes, along which the tangent space of $\varphi(U^{m})$ is constant (i.e., isomorphic to the same linear subspace of $\mathbb{R}^{m}$);
2. With the induced metric, $\varphi(U^{m})$ is flat.


Recall that a submanifold $M$ of a Riemannian manifold $\overline{M}$ is said to be a *full submanifold* if it is not contained in any totally geodesic submanifold $N$ of $\overline{M}$ with $\dim N < \dim \overline{M}$. 


In general, can $\varphi(U^{m})$ be a full submanifold?



Codimension reduction for developable Euclidean submanifold
Fix a generic smooth curve $\gamma$ in $\mathbb{R}^{n+1}$.
The productt $\gamma\times\mathbb{R}^{m-1}\subset\mathbb{R}^{n+m}$ is full, is not it? 


