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Which functions are spherical derivatives?

Mathematics Asked by Giuseppe Negro on December 22, 2020

Let us define the differential operator
$$
Z=x_1 partial_{x_2} – x_2 partial_{x_1}, $$

where $(x_1, x_2, x_3)$ are the standard Cartesian coordinates on $mathbb R^3$. I would like to characterize the functions $hin C^infty(mathbb S^{2})$ such that
$$tag{1}
h=Zf, qquad text{for some }fin C^infty(mathbb S^{2}).$$

The operator $Z$ is one of the generators of the rotation group $SO(3)$, in the sense that
$$tag{2}
Zf(x)=lim_{epsilonto 0}frac1epsilon big( f(R_epsilon x)-f(x)big), $$

where $R_epsilon$ is the matrix
$$
R_epsilon=begin{bmatrix} 
cos epsilon & -sin epsilon & 0 \
sin epsilon & cos epsilon & 0 \
0 & 0 & 1
end{bmatrix}.$$

The formula (2) implies that
$$int_{mathbb S^{2}} Zf(x), dS(x)=0, $$
where $dS$ denotes the standard Lebesgue measure on the sphere. Thus, a necessary condition for (1) to hold is that
$$tag{3}
int_{mathbb S^{2}} h, dS = 0.$$

Also, since $Z$ vanishes at $(0,0, pm 1)$, another necessary condition is
$$
tag{4} h(0,0,pm 1)=0.$$

Are these two last conditions also sufficient?

One Answer

No. A stronger condition is needed. Since the orbints of $Z$ are the circles on constant lattitude, $Zf$ must integrate to zero over each such circle. This can be seen by working in standard spherical coordinates $(theta,varphi):[0,2pi)times(0,pi)to S^2$. In these coordinates, $Z=partial_theta$, so we can apply the fundamential theorem of calculus along the lattitudes, obtaining $$ f(theta,varphi)=f(0,varphi)+int_0^theta Zf(theta',varphi)dtheta' $$ Since $f(0,varphi)=lim_{thetato 2pi^-}f(theta,varphi)$, we must have $int_0^{2pi} Zf(theta',varphi)dtheta'=0$. If you modify your set of conditions, you can check sufficiency by checking if the above expression gives a smooth function on $S^2$ for some choice of $f(0,varphi)$.

Correct answer by Kajelad on December 22, 2020

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