5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ This is probably overkill, but you can take a look at Folland, "Harmonic analysis of the de Rham complex on the sphere. " Crelles 1989 $\endgroup$ – Donu Arapura Mar 8 '11 at 0:26
7
$\begingroup$

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 $\ast f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.