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?

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    $\begingroup$ This is probably overkill, but you can take a look at Folland, "Harmonic analysis of the de Rham complex on the sphere. " Crelles 1989 $\endgroup$ Mar 8, 2011 at 0:26

1 Answer 1


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 $\ast f$.


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