What do the eigenforms of the 1-form Laplace-de Rham operator look like on the 2-sphere, seen as vector fields via the inner product?


For the standard Laplace-de Rham operator on 0-forms (functions) the simple answer is the spherical harmonics. What about for the 1-form operator?



Laplace-deRham operator for 1-forms on the sphere
If $f:S^2\to R$ satisfies $\Delta f=\lambda f$, then 
$$\Delta(df)=(dd^\*+d^\*d)(df)=\lambda df$$
and similarly


$$ \Delta(\ast df)=(dd^\*+d^\*d)(\*df)=\lambda \ast df $$


Since $H^1(S^2)=0$, these are all eigenvectors on 1-forms. Here $\*$ is the Hodge \* operator and $d^\*=-\ast d \ast$.


The vector field is the unique $X$ so that $df(v)=(X,v)$.


On 2-forms all eigenvectors are of the form $\ast f$.


