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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?

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  • $\begingroup$ Off hand, do you have a convenient reference for the "well-known" fact? (To be precise, I know how to prove it, but if it is well-known, the next time I write a paper I may want to just cite it instead!) $\endgroup$ – Willie Wong Aug 8 '18 at 14:17
  • $\begingroup$ @WillieWong: You may check out Vitaly Ushakov, Developable surfaces in Euclidean space, J. Austral. Math. Soc. Ser. A 66 (1999). A more classical reference is Hartman's On isometric immersions in Euclidean space of manifolds with non-negative sectional curvatures $\endgroup$ – MK7 Aug 9 '18 at 16:06
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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?

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