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?

  • $\begingroup$ @Dominik First: who cares that $S^2$ is not open? That's not the point. Second: yes, I want to consider fields defined on open subsets of $\mathbb{R}^3$, but WHAT is true? I gave you two examples going in different directions. $\endgroup$
    – Ivo
    May 2, 2013 at 19:05
  • $\begingroup$ Sorry for wasting your time, I slightly misread your (nice) question! $\endgroup$
    – Dominik
    May 2, 2013 at 19:42
  • $\begingroup$ Observe that you don't need orthogonality for the second vector field -- you just need pointwise linearly independent and nowhere vanishing. Then, since $\mathbb{R}^3$ is oriented, you get a third vector field free. You are therefore trying to find a trivialization of the tangent bundle of the quotient of $U$ by the flow of $X$. In general, this quotient will be a horrible space and won't have a tangent bundle, but maybe something can be said anyway. (Note that in your two examples, the quotients are $\mathbb{R}^2$ are $S^2$ respectively, and in particular very nice.) $\endgroup$
    – Sam Lisi
    May 3, 2013 at 16:38
  • $\begingroup$ @SamLisi Thank you for pointing out that (since $U$ is orientable) linear independece suffices. By the way, as I at first formulated, my question is equivalent to ask when the two-dimensional smooth distribution generated by $X$ (i.e. the subbundle of $TU$ made up of the planes normal to $X$) is trivial as a vector bundle over $U$. $\endgroup$
    – Ivo
    May 4, 2013 at 7:45
  • $\begingroup$ @Yvoz How does this last observation of yours not answer your question? It seems like a characterization of when such a vector field exists. What are you looking for if that isn't the answer? $\endgroup$
    – Sam Lisi
    May 4, 2013 at 19:56

1 Answer 1


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.


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