# Everywhere Normal Vector Field

The title is maybe a little misleading, but here we go.

Not so much a standard problem, but a neat little result. Let $U\subset \mathbb{R}^3$. Let $X$ be a nowhere vanishing vector field on U. Then for each point $p\in U$ there is a surface passing through $p$ such that $X$ is normal to the surface if and only if $\langle X, curl(X)\rangle=0$.

This is nice since it ties back to the Frobenius posts. If $X=(X_1, X_2, X_3)$ in coordinates, then define $\omega=X_1dx+X_2dy +X_3dz$. Some texts use “musical notation,” which is amazingly effective once you get used to it. In which case, we would just say let $\omega=X^\flat$. Now $\omega$ is a smooth 1-form, so it defines a 2-dimensional distribution on $U$ in the standard way $D_p=ker\omega\big|_p$.

Now by definition, $T_pU=D_p\oplus X_p$, so our problem has been rephrased in Frobenius language to say D is integrable iff $\langle X, curl(X)\rangle=0$ since an integral manifold for $D$ through p will satisfy the surface condition.

Thus by the Frobenius Theorem, the problem is reduced to showing $D$ is involutive iff $\langle X, curl(X)\rangle=0$. But D is involutive iff given any two smooth sections Y and Z of D, $d\omega(Y, Z)=0$. In fancy notation, $d\omega=\beta(curl(X))$, in layman’s terms, if $curl(X)=(C_1, C_2, C_3)$, then $d\omega=C_3dx\wedge dy-C_2dx\wedge dz+C_1dy\wedge dz$ (maybe a sign is wrong there, it doesn’t really matter to finish the problem, but I didn’t actually work it out so don’t fully trust it).

So we now have D involutive iff $d\omega(Y, Z)=det(curl(X)|Y|Z)=0$. But this happens iff $curl(X)\in span(Y, Z)$ iff $curl(X)$ has trivial projection onto $X$, which is precisely $\langle X, curl(X) \rangle=0$.