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2
2
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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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6
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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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