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A paper on supersymmetry in 3-dimensions uses results on the spectra of elliptic operators on $S^3$:

The eigenvalues of the vector Laplacian on divergenceless vector fields is $(\ell + 1)^2$ with degeneracy $2\ell(\ell+2)$ with $\ell \in \mathbb{ Z}$.

They came up with a similar statement for spinor fields on the 3-sphere

The eigenvalues of the Dirac operator $D$ is $\pm (\ell + \frac{1}{2})$ with degeneracy $\ell(\ell+1)$ with $\ell \in \mathbb{ Z}$.

On Math.StackExange it was suggested we can restrict the eigenspace decomposition the Laplacian on $SO(4)$ to $S^3$: $$ L^2(SO(4)) = \bigoplus_{\pi \in \mathrm{Irr}[SO(4)]} \pi \otimes \overline{\pi}$$ In our case, how Peter-Weyl Theorem play out when $SO(4)$ acts on bundles of divergenceless vector fields and spinor-fields $S^3$?

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Let $M=G/K$ with $G$ compact and let $(\tau,W_\tau)$ be an irreducible representation of $K$, let $E_\tau$ be the associated $G$-homogeneous vector bundle of $M$. Then, the space of $L^2$-sections of $E_\tau$ decomposes as $$ L^2(E_\tau) = \sum_{\pi\in\widehat G} V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau). $$

The proof follows by Peter-Weyl: $\Gamma^\infty(E_\tau)\simeq C^\infty(G;\tau):=\{f:G\to W_\tau:f(gk)=\tau(k)^{-1}f(g)\}$. For each $\pi\in\widehat G$, we have $V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau) \to C^\infty(G;\tau)$ given by $v\otimes L \mapsto (x\mapsto L(\pi(x^{-1}).v))$. This map is $G$-equivariant.

An other way, $C^\infty(G;\tau) = C^\infty(G,W_\tau)^K \simeq (C^\infty(G)\otimes W_\tau)^K$. Peter-Weyl theorem implies that $$ L^2(E_\tau) = \sum_{\pi\in\widehat G} (V_\pi\otimes V_\pi^*\otimes W_\tau)^K = \sum_{\pi\in\widehat G} V_\pi\otimes (V_\pi^*\otimes W_\tau)^K = \sum_{\pi\in\widehat G} V_\pi\otimes\operatorname{Hom}_K(V_\pi,W_\tau). $$

When $M$ is symmetric (e.g. $M=S^3$) and $G$ semisimple, the Laplace operator acts as the Casimir element of $\mathfrak g$ (the Lie algebra of $G$). Furthermore, the square of the Dirac operator $D^2$ coincides with the Casimir element plus $1/8$ the scalar curvature. See Section 3.5 of T. Friedrich, "Dirac operators in Riemannian geometry".

For example, let $S^{3}=\operatorname{Spin}(4)/\operatorname{Spin}(3)$, let $\tau$ be the spin representation, then $E_\tau$ is the spinor bundle. $\tau$ has highest weight $\frac12 \varepsilon_1$, thus $\operatorname{Hom}_K(V_\pi,W_\tau)\neq0$ if and only if $\pi$ has highest weight $\Lambda = \frac12((2k+1)\varepsilon_1\pm\varepsilon_2)$. The casimir element acts on $\pi_\Lambda$ as $\|\Lambda+\varepsilon_1\|-\|\varepsilon_1\|$ (see Wallach's book, Harmonic analysis on homogeneous vector bundles). Check that is equal to $$ \frac{(2k+1)(2k+5)+1}{4} $$ The sectional curvature is 6, then $D^2$ acts by $(k+3/2)^2$. Finally, the eigenvalues of $D$ are $\pm(k+3/2)$.

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  • $\begingroup$ Still working through it. It looks like you are proving the $L^2$ decomposition using smooth sections of $\Gamma^\infty$ and $C^\infty$. This discussion has almost a functorial nature - at least you can get the spectra for $\nabla^2$ and $D$ in a similar way. $\endgroup$ – john mangual Aug 14 '13 at 12:34
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The spectrum of the Dirac operator on spheres was computed by Christian Bär:

http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.jmsj/1226499694&page=record

See also the following preprint by Fabian Meier for a different approach:

http://arxiv.org/abs/1103.4097

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  • $\begingroup$ Sir, do you know any paper about the eigenvector of Dirac operator on $S^3$? $\endgroup$ – DLIN Dec 30 '19 at 13:02
  • $\begingroup$ Unfortunately not, sorry... $\endgroup$ – Stephan Mescher Dec 30 '19 at 18:33

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