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.

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