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in Sepanski's book Compact Lie groups, he describes the representation theory of SU(2) as being isomorphic to $\mathbb{N}$

(SU(2) acts irreducibly on the (n+1)-dimensional space of homogeneous complex polynomials of degree n on two variables whose basis is $\{z^kw^{n-k}; k = 0, ..., n \}$)

So we have a complete list of the irreducible representations of SU(2). Consider the left regular representation $L:G \rightarrow \mathcal{U}(L^2(SU(2),\mathbb{C}, \mu))$ given by $g\cdot f(x) = f(g^{-1}x)$, and where $\mathcal{U}(L^2(SU(2),\mathbb{C}, \mu))$ is the group of unitary operators on the Hilbert space of square-integrable complex functions relative to the Haar measure on $SU(2)$. By the peter-weyl theorem, this representation can be decomposed as a sum of irreducible finite dimensional subrepresentations of SU(2).

My question is: How do we decompose this left regular representation as a sum of the representations on the space of polynomial functions described on the beggining of the question? More precisely, how do we know which representations appear in the decomposition and with which multiplicity?

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  • $\begingroup$ Great question! $\endgroup$
    – orangeskid
    Commented Sep 20, 2014 at 19:14
  • $\begingroup$ If I remember correctly, the Peter-Weyl theorem also tells you that any irreducible representation $V$ of finite dimension $n$ appears in $L^{2}(G)$ with multiplicity $n$. $\endgroup$ Commented Sep 22, 2014 at 0:10
  • $\begingroup$ They do indeed. I found the answer to this question. $\endgroup$ Commented Sep 22, 2014 at 16:22

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While general theory tells us each irreducible shows up with multiplicity equal to it's dimension, we still may want to know just what functions on $SU(2)$ these correspond to. Rather than describing explicit irreducible factors, it is easier to just describe the isotypic components, but you can find an irreducible just by taking the space generated by any vector inside these isotypic parts.

$SU(2)$ as a manifold is just the three dimensional sphere $S^3$ cut out by the equation $x^2+y^2+z^2+w^2 = 1$ (In fact the multiplication can be taken to just be quaternion multiplication). Sitting inside $L^2(S^3)$ is the space of polynomial functions, this is dense by Stone-Weierstrass and one can check that it is preserved by the action of $SU(2)$. Moreover this action preserves degrees, and you can check that the space of polynomial functions on $S^3$ of degree $d$ has dimension $(d+1)^2$ (Remember that $x^2+y^2+z^2+w^2 - 1 =0$ on this, so we need to account for this). These are exactly the isotypic pieces we are looking for.

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  • $\begingroup$ it took me some time, but I think I understand what you mean. I still can't see $SU(2)$ as $S^3$. Nevertheless I needed an explicit decomposition of the regular representation into irreducible finite-dimensional subspaces. $\endgroup$ Commented Sep 23, 2014 at 12:41
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There's a stronger version of Peter-Weyl theorem, found on this book that states:

All finite-dimensional irreducible representations of $G$ appear in $L$ and also with multiplicity equal to their dimension.

In the case of SU(2), there's exactly one representation of dimension $n$ for every $n \in \mathbb{N}$, so the decomposition takes the form:

$L = \oplus_0^{\infty} V_n^n$.

Where $V_n$ is the space of polynomials with base $\{z^kw^{n-k}; k \leq n\}$ and $V_0$ is the trivial representation.

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