When does a vector field admit orthogonal fields? My question is:

Let $\,X$ be a nonvanishing smooth vector field over an open subset $U \subset \mathbb{R}^3$. Which conditions on $X$ guarantee the existence of a smooth nonvanishing vector field $Y$ on $U$ such that $\,X_p$ is orthogonal to $Y_p$ for each $\,p \in U\,$?

Two examples:


*

*if $\,X$ is a constant field, of course admits (constant) orthogonal fields;

*suppose $\,X$ is the radial field defined by $\,X_p = p\,$ on $\mathbb{R}^3 \smallsetminus \{0\}$ and $\,Y$ orthogonal to $\,X$. Then $\,Y$ is tangent to the sphere $S^2$ and so it must vanish on some point.


Lateral questions could be:

Is the answer related to the topology of normal surface? How?

 A: Let $E$ be the bundle over $U$ whose fibre at each point is the orthogonal complement to $X$. (i.e. $E$ can be identified with $T \mathbb R^3 / \mathbb R X$.) Note that $E$ is a rank $2$ oriented vector bundle over $U$.
For notational simplicity, take $X$ from now on to have norm 1. We may then see $X$ as a map $X \colon U \to S^2$. Furthermore, any such map can be reinterpreted as giving us an oriented rank 2 vector bundle of the kind you describe. Two homotopic maps clearly give isomorphic bundles.
Your question can be reformulated as wanting to find a nowhere vanishing section of $E$, i.e. you want to know when the Euler class $e \in H^2(U, \mathbb Z)$ of $E$ vanishes. If we have a 2-cycle in $U$ represented by the image of a map $f \colon S \to U$ for a surface $S$ (and by Thom these are actually all of them, at least if $U$ is a CW-complex... not sure how bad $U$ is allowed to be for this to hold), then we can compute the pairing of $<e, f_* [S]>$ by the degree of the map $X\circ f \colon S \to S^2$.
In particular, if $X$ is homotopic in the maps $U \to S^2$ to a constant map, we are done. I suspect the converse is true too (i.e. that vanishing Euler class means that $X$ must be homotopic through non-vanishing vector fields to the constant vector field), at least if $U$ is nice enough.
