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
$$\bigl\langle \frac{\partial \xi}{\partial u\_{1}}, X\_{j}\bigr \rangle = \dotsb = \bigl\langle \frac{\partial \xi}{\partial u\_{m-k}}, X\_{j}\bigr \rangle =0 \quad \text{for all $j=1,\dotsc, k$}?$$



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


