Vtyjkyuk

Consider a $C^1$ map $f:R^2to R^3$ such that $df_x:R^2to R^3$ has rank $2$ everywhere. Define the tangent plane $T_x$ as $df_x(R^2)subset R^3$. Suppose a vector field $X$ in $R^3$ is orthogonal to $T_x$ for all $x$, i.e., $X(f(x))cdot Y=0$ for all $xin R^2$ and $Yin T_x$. Show that $Xcdot (nabla times X)=0$ at all points $f(x)$.

My thoughts: any vector field on $R^3$ can be written as $a(p,q,r)partial_p+b(p,q,r)partial q+c(p,q,r)partial _r$ where $p,q,r$ are coordinates in $R^3$. So $$nablatimes X=(partial_qcpartial_r-partial_rbpartial_q,partial_rapartial_p-partial_pcpartial_r,partial_pbpartial_q-partial_qapartial_p)$$ and

$$Xcdot (nabla times X)=apartial_ppartial_qcpartial_r-apartial_ppartial_rbpartial_q+bpartial_qpartial_rapartial_p-bpartial_qpartial_pcpartial_r+cpartial_rpartial_pbpartial_q-cpartial_rpartial_qapartial _p$$
I need to show that the above expression applied to $f(x)$ is zero for all $xin R^2$. Further, $$X(f(x))=(apartial_pf(x),bpartial_qf(x),cpartial_rf(x))\Y=(partial_xf_1u+partial_yf_1 v,partial_xf_2u+partial_yf_2 v,partial_xf_3u+partial_yf_3 v)$$ where $f=(f_1,f_2,f_3), u,v,in R$. And we are given that $X(f(x))cdot Y=0$. Even if I expand this, I don't know what to do next...

Differential forms for the win.

Think of the image of $f$ locally as an (embedded) surface $M$. Take a nonzero $1$-form $omega$ on an open subset of $Bbb R^3$ so that $M$ is an integral manifold of $omega=0$. [For example, express $M$ as a level set of a smooth function $g$ and take $omega = dg$.] The fact that $omega=0$ is integrable tells you that $domega wedgeomega = 0$.

Under the usual dictionary between $1$-forms and vector fields in $Bbb R^3$, if $X$ corresponds to $omega$, $textcurl, X = nablatimes X$ corresponds to the $2$-form $domega$, and the dot product $Xcdot(nablatimes X)$ corresponds to the $3$-form $omegawedge domega$. (It is easy to check, in coordinates, that if you have vector fields $X,Y$ corresponding to $1$-forms $omega,psi$, then $Xcdot Y = star(omegawedgestarpsi)$.)

By the way, note that if we replace $X$ by any functional multiple $phi X$, $(phi X)cdot(nablatimes phi X) = phi^2 Xcdot(nablatimes X) + phi Xcdot (nablaphitimes X) = 0$, as well. From the differential forms viewpoint, we're looking at $(phiomega)wedge d(phiomega) = phi^2 omegawedge domega + phiomegawedge dphiwedgeomega$, and $omegawedge dphiwedgeomega = dphiwedgeomegawedgeomega = 0$, since $omega$ is a $1$-form.

EDIT: More interesting, if we replace $X$ by $X+Z$ for any vector field $Z$ vanishing on $M$, why does $(X+Z)cdotnablatimes(X+Z)$ still vanish on $M$? This boils down to seeing that $Xcdotnablatimes Z = 0$ on $M$. This is true the component of the curl orthogonal to $M$ comes only from partial derivatives of components of $Z$ in directions tangent to $M$, and these vanish because $Z=0$ on $M$. [This, too, is easier to see using differential forms, since we don't need to worry about a particular coordinate system. Choose local coordinates $(x,y,z)$ so that $M = z=0$. Taking $omega = dz + eta$, where $eta = P,dx+Q,dy+R,dz$ vanishes identically on $M$, we want to see that $domegawedgeomega$ still vanishes on $M$. Well, $detawedge dz = left(fracpartial Qpartial x-fracpartial Ppartial yright) dxwedge dywedge dz$, and these partials vanish along $z=0$ since the functions are identically $0$ on $z=0$.]

If you are not familiar with differential forms and integrability, I would recommend proceeding as follows. Write $M$ as $g=0$ and take $X = nabla g$. (Recall that gradients are orthogonal to level sets.) Then $nablatimes X = 0$ and we're done. From my earlier comment, rescaling $X$ won't change the result of the computation.

If you want to justify the assertion that $M$ is (locally) a level set of a smooth function, you need something like the implicit function theorem. It is locally a graph over one of the coordinate planes, say $z=psi(x,y)$. Then take $g(x,y,z) = z-psi(x,y)$.