4
$\begingroup$

Deitmar and Echterhoff write in their book Principles of Harmonic Analysis that `It follows from the Peter-Weyl theorem that the $SU(2)$ representation on $L^2(S^3)$ is isomorphic to the orthogonal sum $\oplus_{m\ge 0}(m+1)P_m$ where $P_m$ is the space of homogeneous polynomials of degree $m$.'

They provide no further explanation of this statement, so here is my attempt at understanding it: we can use the homeomorphism $SU(2)\simeq S^3$ and let $SU(2)$ act on $L^2(S^3)\simeq L^2(SU(2))$ by $$(\pi(g)f)(x)=f(xg).$$ Since the group acting, namely $SU(2)$ is compact the Peter-Weyl theorem says that $L^2(S^3)$ decomposes as a hilbert space direct sum of irreducible representations, namely $(\pi_m, P_m)$ each appearing with multiplicity $\dim \text{Hom} (\pi_m, \pi)$. So it remains to prove that $\dim \text{Hom} (\pi_m, \pi)=m+1.$

Now my first question is whether computing the dimension of the space of intertwiners is elementary.

Second, are there any concrete applications for harmonic analysis on $S^3$ that this result might be helpful for?

Lastly, how can one come to grips with the decomposition of $L^2(S^n)$ for $n>3$?

Any (partial) answers to any of these questions is highly appreciated.

$\endgroup$
5
$\begingroup$

1) Your formula for the decomposition of the regular representation is not the whole statement of the Peter-Weyl theorem. It is part of the Peter-Weyl theorem that the multiplicity $\dim(Hom(\pi_m,\pi))$ in the regular representation equals the dimension of the representation $\dim(\pi_m)$ which is of course $(m+1)$.

2) Computation and properties of spherical harmonics is the most classical applications, and most others stem from that. See below.

3) In that case $S^n$ is not itself isomorphic to a Lie group; however, it is still a symmetric space $\textrm{SO}(n)/\textrm{SO}(n-1)$ and we can still use Peter-Weyl theory, only in a more delicate way. Spherical harmonics arise here as well; see http://www.springer.com/cda/content/document/cda_downloaddocument/9781461479710-c1.pdf?SGWID=0-0-45-1445132-p175259457 for an introduction to these things.

$\endgroup$
  • $\begingroup$ Thanks for the link. I'll have a look at it soon. I am a bit confused about the computation of multiplicity. How do we know that the representation $\pi_m$ appears with multiplicity $m+1$ in the decomposition of $L^2(S^3)$? I think a calculation is needed here. For instance if we replace $S^3$ with $S^2$ then the calculation can be done with Frobenius reciprocity and the final result is that $\pi_m$ appears with multiplicity 0 or 1 depending on whether $m$ is odd or even. $\endgroup$ – EPS Jan 1 '15 at 5:42
  • $\begingroup$ @Sam a lot of calculation is needed, not a single one. It uses the orthogonality of characters and is, as I said, part of the Peter-Weyl theorem. Edit: it does not need Frobenius reciprocity. $\endgroup$ – guest Jan 1 '15 at 5:45
3
$\begingroup$

Peter-Weyl says a bit more than even what guest states: it tells you how $L^2(G)$ decomposes under the natural $G \times G$ action given by multiplication by $g \in G$ on the left and multiplication by $g^{-1} \in G$ on the right. This decomposition is as a Hilbert space direct sum

$$\bigoplus_V V \boxtimes V^{\ast}$$

where $\boxtimes$ denotes the external tensor product and the direct sum runs over all irreducible representations of $G$. This is exactly what one expects from the finite case, since in the finite case $\mathbb{C}[G]$ is naturally isomorphic to the direct sum $\bigoplus_V \text{End}(V)$ of the endomorphism algebras of the irreducible representations of $G$. Restricting to one copy of $G$ gives the desired multiplicity result, that $V$ appears with multiplicity $\dim V^{\ast} = \dim V$.

$\endgroup$

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.