Parametrization of $k$-ruled submanifold: can we choose the base to be orthogonal to the rulings?

Question

A smooth $m$-dimensional submanifold of $\mathbb{R}^{d}$ is said to be $k$-ruled if it is foliated by $k$-dimensional planes, called rulings.

Let $M$ be a $k$-ruled submanifold. Then $M$ can be parametrized (locally) by a smooth map $\sigma \colon (U\subset \mathbb{R}^{m-k}) \times \mathbb{R}^{k} \to \mathbb{R}^{d}$ given by $$\sigma(u_{1}, \dotsc, u_{m-k}, v_{1}, \dotsc, v_{k}) = \xi(u_{1}, \dotsc, u_{m-k}) +\sum_{j=1}^{k} v_{j}X_{j}(u_{1}, \dotsc, u_{m-k}),$$ where $\xi$ is a "base" submanifold and $(X_{1}, \dotsc, X_{k})$ an orthonormal frame along $\xi$.

In particular, if $k=m-1$, then we can always choose the curve $\xi$ to be orthogonal to all the $(m-1)$-planes, i.e., such that $$\langle \xi', X_{j} \rangle =0 \quad \text{for all $j=1,\dotsc, m-1$}.$$

I wonder if the same holds in general.

Question. Does there exist a parametrization of $M$ such that $$\bigl\langle \frac{\partial \xi}{\partial u_{1}}, X_{j}\bigr \rangle = \dotsb = \bigl\langle \frac{\partial \xi}{\partial u_{m-k}}, X_{j}\bigr \rangle =0 \quad \text{for all $j=1,\dotsc, k$}?$$

Answer 1

The answer is already 'no' in the first nontrivial case: A 3-manifold in $\mathbb{R}^d$ (where $d>3$) that is ruled by lines (i.e., $k=1$).

One can see this as follows: As the OP notes, one can write $\sigma$ as above in the form $$ \sigma(u_1,u_2,v_1) = \xi(u_1,u_2) + v_1 X(u_1,u_2) $$ where, without loss of generality, one can assume that $X\cdot X = 1$ and that the three $\mathbb{R}^d$-valued mappings $X,\ \partial_{u_1}\xi,\ \partial_{u_2}\xi$ are everywhere linearly independent. It's easy to see that a surface $S$ in the domain of $\sigma$ will map via $\sigma$ to a surface in $\mathbb{R}^d$ that is orthogonal to the $X$-ruling if and only if the surface is an integral surface of the $1$-form $$ \theta = X\cdot \mathrm{d}\sigma = \mathrm{d} v_1 + (X\cdot \partial_{u_1}\xi)\,\mathrm{d}u_1 + (X\cdot \partial_{u_2}\xi)\,\mathrm{d}u_2\,. $$ (Note the use of $X\cdot X = 1$, which implies $X\cdot \mathrm{d}X = 0$.) In particular, the $2$-form $$ \mathrm{d}\theta = \bigl((\partial_{u_1}X\cdot \partial_{u_2}\xi)-(\partial_{u_2}X\cdot \partial_{u_1}\xi)\bigr)\,\mathrm{d}u_1\wedge \mathrm{d}u_2 $$ must vanish on such a surface.

However, it's easy to write down specific examples for which the function $\bigl((\partial_{u_1}X\cdot \partial_{u_2}\xi)-(\partial_{u_2}X\cdot \partial_{u_1}\xi)\bigr)$ is nowhere vanishing. In such examples, no such surface $S$ can exist because the vanishing of $\theta$ and $\mathrm{d}u_1\wedge \mathrm{d}u_2$ on $S$ implies that the three differentials $\mathrm{d}u_1$, $\mathrm{d}u_2$, and $\mathrm{d}v_1$ satisfy two independent linear relations on $S$, which is impossible.