TITLE: Laplace-deRham operator for 1-forms on the sphere
QUESTION [5 upvotes]: 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?

REPLY [8 votes]: 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 $\ast f$.