Introduction andPrerequisites

Representation theory is important in various branches of mathematicsand physics. When studying representation of finite groups, we havequite some algebra and combinatorics. When differentiation (moreprecisely, smoothness) joins the party, we have Lie group, involvingcalculus, linear algebra, geometry and much more. Especially, theoriesaround $SU(2)$ and $SO(3)$ are of great importance. On onehand, they are those simplest non-elementary and high-dimensional Liegroups. On the other hand, they describes rotations in $\mathbb{C}^2$ and $\mathbb{R}^3$ respectively, which is"physically realistic". I believe students in physics have more tosay.

In this post we develop a way to study irreducible representations ofthese two Lie groups, in a mathematician's way. I try my best to makesure that everything is down-to-earth, and everything can be "reduced"to 19th (pre-modern) mathematics.

Nevertheless, the reader has to be assumed to be familiar withelementary languages of representation theory (and you know that, thereare a lot of abuse of language), which I think is not a problem becauseotherwise you wouldn't be reading this post. You need to recalleigenvalue theories in linear algebra, as well as Fourier series. Weneed the fact that the trigonometric system is complete. In other wordstrigonometric polynomials are dense in the space of continuousfunctions. $\def\sym{\operatorname{Sym}}$

We will first study $SU(2)$ and afirst classification of irreducible representations of $SO(3)$ follows at once. This is because wehave an isomorphism

\[SU(2)/\{-I,I\} \cong SO(3).\]

This is to say, $SU(2)$ is a"double cover" of $SO(3)$. To seethis, notice that $SU(2) \cong S^3$and $SO(3) \cong \mathbb{R}P^3$ asLie groups, meanwhile $\mathbb{R}P^3 \congS^3/\{-1,1\}$ can be considered as the definition.

Of course, by representation we mean finite dimensional and unitaryrepresentations.

IrreducibleRepresentations of the Special Unitary Group

Indeed it seems we have nowhere to start. Instead of trying to findall of them, we will try to work on seemingly immediate representationsand it turns out that they are all we are looking for.

Let $V_0$ be the trivialrepresentation on $\mathbb{C}$ and$V_1$ be the standard representationon $\mathbb{C}^2$, which is given byordinary matrix multiplication. These representations are irreducible.We want to extend this family to $V_n$ for $n \ge2$. It is natural to think about generate representations ofhigher dimensions through $V_1$. Hereare several ways available.

Direct sum: $\bigoplus_{i=1}^{n}V_1$. The dimension is$2n$ and unfortunately, therepresentation is determined by each component so essentially there isno "new thing".
Tensor product: $\bigotimes_{i=1}^{n}V_1$. The dimension is$2^n$ which is way too big.
Wedge product: $\bigwedge^{n}V_1$. It stops at $n=2$ and we have to deal with $u \wedge v = - v \wedge u$. This can beannoying.
Symmetric product: $\sym^{n}V_1$. The dimension is $n+1$ and it doesn't stop. Besides, it canbe understood as homogeneous polynomials of degree $n$ in two variables. This is a fantasticchoice. Besides we have $\sym^0V_1=V_0$ so nothing is abruptly excluded.

Spaces of HomogeneousPolynomials

Put $V_n=\sym^nV_1$, which can beunderstood as the space of homogeneous polynomials of degree $n$ in variables $z_1$ and $z_2$. $V_n$ therefore has a canonical basis

\[P_k=z_1^k z_2^{n-k}.\]

And we will make use of it later.

Definition of therepresentation

For each $g \in SU(2)$, we have aleft action

\[\begin{aligned}\rho:SU(2) &\to \operatorname{Aut}(V_n), \\ g &\mapsto (P(z) \mapsto P(zg)).\end{aligned}\]

In other words, $\rho(g)P(z)=P(zg)$ where $z=(z_1,z_2)$ and $zg$ is matrix multiplication. Each $g \in SU(2)$ has matrix representation

\[g=\begin{pmatrix}\alpha & \beta \\-\overline{\beta} & \overline{\alpha}\end{pmatrix}, \quad |\alpha|^2+|\beta|^2=1.\]

Then

\[zg=(\alpha z_1-\overline{\beta}z_2,\beta z_1+\overline{\alpha}z_2)\]

When there is no confusion, we will write $gP(z)=P(zg)$, viewing $g$ itself as an automorphism of $V_n$. One can also replace $SU(2)$ with $GL(2,\mathbb{C})$ but we are not studyingthat bigger one.

Since $z \mapsto zg$ is ahomogeneous map of degree $1$ as itis linear and is non-degenerate, we have $gP(z) \in V_n$. In other words, $V_n$ are $SU(2)$-invariant. We now have awell-defined representation. Note $V_0=\mathbb{C}$ so the representation istrivial, and $V_1=\mathbb{C}^2$yields linear maps. Again, nothing is abruptly excluded. Even moresatisfyingly, those $V_n$ are allirreducible.

Irreducibility

Proposition 1. The representations $V_n$ are irreducible.

Proof. By Schur's lemma, we need to show that each $SU(2)$-equivariant automorphism $A$ of $V_n$ is a non-zero multiple of theidentity, i.e. $A=\lambda I$ for some$\lambda \ne 0$. By definition, foreach $g \in SU(2)$, we have $A\rho(g)P=\rho(g)AP$ for all $P \in V_n$. And for simplicity we write$Ag=gA$, realising $g$ as a linear transform of $V_n$, instead of an element of $SU(2)$.

The group $SU(2)$ can becomplicated, but $U(1) \cong S^1$ issimple and can be considered as a subgroup of $SU(2)$ in two ways. We show that these twoways are just enough to expose the irreducibility of $V_n$.

First of all we embed $S^1$ into$SU(2)$ by

\[a \mapsto \begin{pmatrix}a & 0 \\ 0 & a^{-1} \end{pmatrix}.\]

Call the matrix right hand side $g_a$. Then

\[g_a P_k=(az_1)^{k}(a^{-1}z_2)^{n-k}=a^{2k-n}z_1^kz_2^{n-k}=a^{2k-n}P_k\]

for all $k$. This is to say, $P_k$ is the eigenvectorcorresponding to eigenvalue $a^{2k-n}$. As $g_aA=Ag_a$, information on eigenvalues andeigenvectors can help a lot so we dig into it first.

Since $\{P_k\}$ are linearlyindependent, under this basis, we have a matrix representation

\[\rho(g_a) = \operatorname{diag}(a^{-n},a^{-n+2},\dots,a^{n-2},a^n).\]

but we don't know how eigenspaces are spanned because we may have$a^j=a^k$ for $j \ne k$. However, the number $a$ can always be chosen that $a^{-n},a^{-n+2},\dots,a^n$ are pairwisedistinct (for example, one can pick $a$ to be a primitive $m$-th root of $1$ and $m$ is big enough). As a result, $g_a$ has $n$ distinct eigenvalues. Therefore, the$a^{2k-n}$-eigenspace can only begenerated by $P_k$.

On the other hand, by definition of $A$, we have

\[Ag_a P_k = g_aAP_k = A a^{2k-n}P_k = a^{2k-n}AP_k.\]

Hence $AP_k$ lies in $a^{2k-n}$-eigenspace. Therefore we have$AP_k=c_kP_k$ for some $c_k \ne 0$. In other words, $P_k$ is the $c_k$-eigenvector of $A$. We obtain another matrixrepresentation under the basis $\{P_k\}$

\[A = \begin{pmatrix}c_1 & & \\ & \ddots & \\ & & c_n\end{pmatrix}.\]

We want this matrix to be a scalar matrix. The result follows fromanother embedding of $U(1)$ into$SU(2)$. Note $a \in S^1$ can be determined by $t \in [0,2\pi)$, and we therefore have amatrix

\[g_t = \begin{pmatrix}\cos{t} & -\sin{t} \\\sin{t} & \cos{t}\end{pmatrix} \in SU(2).\]

Still we have $Ag_t=g_tA$. As wecan see,

\[\begin{aligned}Ag_tP_n &= A(z_1\cos{t}+z_2\sin{t})^n \\ &= A\sum_{k=0}^{n}{n \choose k}(z_1\cos{t})^k(z_2\sin{t})^{n-k} \\ &= A\sum_{k=0}^{n}{n \choosek}(\cos{t})^k(\sin{t})^{n-k}z_1^k z_2^{n-k} \\ &= A\sum_{k=0}^{n}{n \choose k}(\cos{t})^k(\sin{t})^{n-k}P_k \\ &= \sum_{k=0}^{n}{n \choose k}(\cos{t})^k(\sin{t})^{n-k}AP_k \\ &= \sum_{k=0}^{n}{n \choose k}(\cos{t})^k(\sin{t})^{n-k}c_kP_k.\end{aligned}\]

This follows from our observation on eigenvalues. Next, weimmediately use the eigenvalue $c_n$to obtain

\[g_t AP_n = g_t c_nP_n = c_n \sum_{k=0}^{n}{n \choosek}(\cos{t})^k(\sin{t})^{n-k} P_k.\]

This is the definition of $g_tP_n$. Comparing coefficients of $P_k$, we must have $c_k=c_n$ for all $0 \le k \le n$. Recall that $\{P_k\}$ is a basis so coefficients mustbe unique for a given vector. But we have already obtained what we want:$A=c_n I$. $\square$

Characters and FourierTransform

So far we have used diagonalisation of representations of $SU(2)$ but the diagonalisation of $SU(2)$ itself is not touched yet. Neitherhave we made use of character functions. So now we invite them to theparty.

Let's recall diagonalisation in $SU(2)$. Pick $g\in SU(2)$. First of all it is diagonalisable. Let $\lambda_1$ and $\lambda_2$ be their two eigenvalues, then$|g|=\lambda_1\lambda_2=1$. Thereforewe have

\[g \sim \begin{pmatrix}\lambda & 0 \\ 0 & \lambda^{-1}\end{pmatrix} \sim \begin{pmatrix}\lambda^{-1} & 0 \\ 0 &\lambda \end{pmatrix}\]

where $\lambda$ is one of theeigenvalues of $g$. Since thediagonalised matrix is still in $SU(2)$, we have $|\lambda|=1$, i.e., $\lambda \in S^1$. We therefore write $g \sim e(t) \sim e(-t)$ where

\[e(t) = \begin{pmatrix} \exp(it) & 0 \\ 0 & \exp(-it)\end{pmatrix}.\]

We see, $e(s) \sim e(t)$ if andonly if $s = \pm t \mod 2\pi$. Byperiodicity of $\exp$ function, wealso see $e(t)$ is in particular$2\pi$-periodic. If $f:SU(2) \to \mathbb{C}$ is a classfunction, then $f \circ e:\mathbb{R} \to\mathbb{C}$ is an even $2\pi$-periodic function. Conversely, givenan even $2\pi$-periodic function$h:\mathbb{R} \to \mathbb{C}$, we canrecover it as a class function, and the process is as follows.

Define $\Lambda:SU(2) \to S^1$sending $g \in SU(2)$ to theeigenvalue of $g$ with non-negativeimaginary part (one can also pick non-positive one, because $h$ is even). Then $E:SU(2) \to [0,\pi]$ given by $g \mapsto \frac{1}{i}\log\Lambda(g)$ is awell defined function sending $g$into $\mathbb{R}$ and $h \circ E:SU(2) \to \mathbb{C}$ is a classfunction. Besides we have $E \circ e(t)= \pmt \mod 2\pi$ and $e \circE(g)$ is the diagonalisation of $g$. Therefore $h \circ E \circ e(t)=h(t)$ and $f \circ e \circ E(g)=f$ as isexpected.

With help of this $e(t)$ and $E(t)$, we have this correspondence

\[\{\text{Class functions }SU(2) \to \mathbb{C}\} \longleftrightarrow\{\text{even }2\pi-\text{periodic function }\mathbb{R} \to \mathbb{C}\}\]

Recall that the space on the right hand side has a countable uniformbasis

\[1,\cos{t},\cos{2t},\dots.\]

In other words, $\{\cos{nt}\}_{n \ge0}$ spans a dense subspace. This is about the completeness oftrigonometric system. Since there are only even functions, $\sin{nt}$ are excluded. For a reference tothe completeness, one can check 4.25 Real and Complex Analysisby W. Rudin.

For class functions, we certainly want to know about characters. Let$\chi_n$ be the character of $V_n$, then

\[\chi_n(e(t))=\operatorname{tr}(\rho(e(t)))=\operatorname{tr}(\operatorname{diag}(\exp(it)^{-n},\dots,\exp(it)^n))=\sum_{k=0}^{n}e^{i(n-2k)t}.\]

When $t \in \pi\mathbb{Z}$, then$\chi_n(e(t)) \in \mathbb{Z}$.Otherwise, as a classic exercise in calculus, we have

\[\kappa_n(t)=\chi_n(e(t))=\frac{\sin(n+1)t}{\sin{t}}.\]

We have $\kappa_0(t)=1$. For $\kappa_n(t)$ when $n >0$, we have

\[\kappa_n(t)=\frac{\cos{nt}\sin{t}+\sin{nt}\cos{t}}{\sin{t}}=\cos{nt}+\kappa_{n-1}(t)\cos{t}.\]

We see $\kappa_1(t)=2\cos{t}$. Byinduction, every $\kappa_n(t)$ is apolynomial in variables $1,\cos{t},\dots,\cos{nt}$. Therefore $\{\kappa_n(t)\}_{n \ge 0}$ spans the samespace as $\{\cos{nt}\}_{n \ge 0}$,which is dense in the space of even $2\pi$-periodic functions. Note the $\kappa_n(t)$ are linearly independent,because the leading term is $\cos{nt}$.

The argument above shows that $\chi_n$ spans a dense subspace in thespace of class functions. In other word, $\chi_n$ is the Fourier basis of classfunctions. As we all know, Fourier series is powerful. Let's see howpowerful it is in the calculus of Lie group $SU(2)$ itself.

Proposition 2. For continuous class function $f:SU(2) \to \mathbb{C}$, we have
\[\int_{SU(2)}f(x)dx = \frac{1}{\pi}\int_0^\pi f \circ e(t)\sin^2{t}dt.\]

Proof. On one hand, since the $V_n$ are irreducible, by fixed pointtheorem of representations,

\[\int_{SU(2)}\chi_n(x)dx = \dim V_n^{SU(2)} = \begin{cases} 1 & n=0,\\ 0 & n>0. \end{cases}\]

Here, for a group $G$ and arepresentation $V$, $V^G$ is the fixed point set, i.e. thespace of elements that are fixed by the action of $G$ on $V$. Since $\chi_n$ is irreducible, fixed points canonly be $0$ unless the representationitself is trivial. Now we move on and check the right hand side.

On the right hand side we are looking for even $2\pi$-periodic continuous functions,reflecting the denseness of $\kappa_n(t)$. However we have $\int_{-\pi}^{\pi}\kappa_1(t)dt=\pi$ so itdoes not vanish on $n>0$. However,if we multiply it by $\sin^2{t}$,then it is transformed into the form $\sin{mt}\sin{nt}$ and we are familiar withthis orthonormality. More precisely,

\[\frac{1}{2\pi}\int_{-\pi}^{\pi}f\circ e(t)\sin^2{t}dt =\frac{1}{\pi}\int_{-\pi}^{\pi}\sin(n+1)t\sin{t}dt = \begin{cases} 1& n=0, \\ 0 & n>0. \end{cases}\]

Since the functional $h \mapsto\frac{1}{2\pi}\int_{-\pi}^{\pi}h\sin^2{t}dt$ is continuous inthe uniform topology and $\kappa_n$spans a dense subspace, the result is now obtained. $\square$

Finally, surprisingly and satisfyingly enough, the denseness haveactually axed out all other possibilities of irreducible representation.In other words, our search in symmetric products is optimal. We can seethis through Parseval's identity. This is the heart of this blogpost.

Proposition 3. Every irreducible representation of$SU(2)$ is isomorphic to one of the$V_n$.

Proof. Suppose we have a character that is different fromall of the $\chi_n$. Then theorthonormality shows that $\langle\chi,\chi_n \rangle = 0$ for all $n\ge 0$ and $\langle \chi,\chi\rangle=1$. Now let's see why this is absurd.

Since $\{\chi_n\}_{n \ge 0}$ spansa dense subspace in the space of class functions, we actually have

\[\chi = \sum_{n = 0 }^{\infty} a_n \chi_n.\]

Therefore

\[\langle \chi,\chi_n \rangle = \int_{SU(2)}\overline\chi(x)\chi_n(x)dx =a_n=0,\quad n \ge 0\]

and

\[\langle \chi,\chi \rangle = \sum_{n=0}^{\infty}\langlea_n\chi_n,a_n\chi_n \rangle = \sum_{n=0}^{\infty}|a_n|^2=1.\]

It is impossible to have the sum of $0$ to be $1$. $\square$

IrreducibleRepresentations of the Special Orthonormal Group (FirstClassification)

Now we head to $SO(3)$. In factthe result follows immediately from the surjection

\[\pi:SU(2) \to SO(3).\]

We have $\ker\pi=\{-I,I\}$. Let$W$ be a representation of $SO(3)$, i.e., we have a map

\[\rho:SO(3) \to GL(W).\]

Then

\[\pi^\ast\rho:SU(2) \to GL(W)\]

by $g \mapsto \rho(\pi(g))$ is aninduced representation, and we write $\pi^\ast W$. If $W$ is irreducible, then $\pi^\ast W$ is also irreducible. Inparticular, $\pi^\ast\rho(-I)=\operatorname{id}_W$.

On the other hand, if $\vartheta:SU(2) \toGL(V)$ is an irreducible representation where $\vartheta(-I)=\operatorname{id}_V$, thenwe have an associated representation

\[\pi_\ast\vartheta:SO(3) \cong SU(2)/\{I,-I\} \to GL(V)\]

given by $g\ker\pi \mapsto\vartheta(g)$. Let's denote it by $\pi_\ast V$. Again, if $V$ is irreducible, then $\pi_\ast V$ is irreducible.

Therefore we have realised a correspondence

\[\{\text{Irreducible representations of $SO(3)$}\} \\\updownarrow \\\{\text{Irreducible representations of $SU(2)$ where $-I$ acts asidentity.}\}\]

So it remains to determine those of $SU(2)$. Let $\rho_n:SU(2) \to GL(V_n)$ be anirreducible representation, then

\[\rho_n(-I)P(z)=P(z(-I))=P(-z)=(-1)^nP(z)\]

because $P \in\mathbb{C}[z_1,z_2]$ is homogeneous of degree $n$. Therefore $-I$ acts as an identity if and only if$n$ is even. We obtain

Proposition 4. Every irreducible representation of$SO(3)$ is of the form
\[W_n = \pi_\ast V_{2n}\]
where $V_{2n}$ is described inproposition 2.

This is, of course, just a first classification. But to introduce aclassification as explicit as what we have done for $SU(2)$, there has to be another post. As aquick overview, here is the result.

Let $P_{\ell}$ be the complexvector space of homogeneous polynomials in three variables of degree$\ell$, which can be considered asfunctions on $\mathbb{R}^3$immediately. This setting makes sense immediately, just as what we havedone for $SU(2)$. Then, in fact,

\[W_\ell=\mathfrak{H}_\ell = \{f \in P_\ell:\Delta f=0\}.\]

This is to say, $W_\ell$ can beunderstood as harmonic homogeneous polynomials in $\mathbb{R}^3$, which can also beconsidered to be uniquely determined on the unit sphere $S^2$.

Reference

Tendor Bröker and Tammo tom Dieck, Representations of CompactLie Groups.
Walter Rudin, Real and Complex Analysis, 3rd Edition.