# Introduction

In our previous post on the irreducible representations of $SU(2)$ and $SO(3)$, the irreducible representations of $SU(2)$ has been determined explicitly: $V_n=\operatorname{Sym}^n\mathbb{C}^2$, and irreducible representations $W_n$ of $SO(3)$ correspond to $V_{2n}$.

The result is satisfying for $SU(2)$ but not for $SO(3)$. We hope it has something to do with $\mathbb{R}^3$ but $V_{2n}$ has not. In this post, we are delivering a much clear characterisation of $W_n$.

This post would be relatively easier to read. Other than the basic language of representation theory (of Lie groups), only multivariable calculus is needed.

## General Strategy & Playground

Like in the previous post, we first determine a good playground and then show that this is all we need. The playground here is

$P_\ell=\mathbb{C} \otimes_\mathbb{R}\operatorname{Sym}^\ell \mathbb{R}^3.$

The reason for the symmetric product of $\mathbb{R}^3$ is simple: we will be working on homogeneous polynomials. We complexified this space to make sure that we will not worry about eigenvalues (of $SO(3)$). In other words, $P_\ell$ is the complex vector space of homogeneous polynomials in three variables, viewed as functions on $\mathbb{R}^3$.

Recall that

$\dim \operatorname{Sym}^\ell \mathbb{R}^3 = {\ell+3-1 \choose \ell}={\ell + 2 \choose \ell}={\ell+2 \choose 2}=\frac{(\ell+2)(\ell+1)}{2}.$

Therefore $\dim P_\ell=\frac{(\ell+2)(\ell+1)}{2}$, as a $\mathbb{C}$-vector space.

We will extract what we want from the spaces of this form.

# The Representation of SO(3) And the Laplacian

The action of $SO(3)$ or $GL(3,\mathbb{R})$ in general on $P_\ell$ is defined in a similar way. For any $A \in GL(3,\mathbb{R})$ and $f \in P_\ell$, we define

$(Af)(x)=f(xA).$

Here, $x=(x_1,x_2,x_3) \in \mathbb{R} \times \mathbb{R} \times \mathbb{R}$, and $xA$ is a product of $x$ and $A$ in sense of matrix multiplication. It is easy to verify that this indeed gives rise to a group representation.

To study this representation, we need to find some morphisms $P_\ell \to P_\ell$. The most obvious choice is the Laplacian, which is given by

$\Delta:f \mapsto \left(\frac{\partial^2}{\partial x_1^2}+\frac{\partial^2}{\partial x_2^2}+\frac{\partial^2}{\partial x_3^2}\right)f.$

In other words, $\Delta$ is the trace of the Hessian matrix of $f$. Trace is used in representation theory to define character so there is a chance to find its good connection to the representation of $SO(3)$.

We shall also not forget the kernel of the Laplacian, which is called harmonic polynomials of degree $\ell$ in this context:

$\mathfrak{H}_\ell = \{f \in P_\ell:\Delta{f}=0\}.$

Since functions in $P_\ell$ are homogeneous, the value of $f$ at a point $x$ is determined by the value at $\frac{x}{\|x\|} \in S^2$, the unit sphere. Therefore we also call $\mathfrak{H}_\ell$ the spherical harmonics of degree $\ell$. And we certainly need to know the nullity of $\Delta$.

Lemma 1. The dimension of $\mathfrak{H}_\ell$ is $2\ell+1$.

Proof. First of all we perform a Taylor expansion of $f \in P_\ell$ with respect to the first variable $x_1$:

$f(x_1,x_2,x_3)=\sum_{k=0}^{\ell}\frac{f_k(x_2,x_3)}{k!}x_1^k.$

Here, $f_k(x_2,x_3)$ is homogeneous of degree $\ell-k$ in $x_2,x_3$. Therefore we only need to study one term of the right hand side.

\begin{aligned} \Delta \frac{f_k(x_2,x_3)}{k!}x_1^k &= \frac{f_k(x_2,x_3)}{k!}k(k-1)x_1^{k-2}+\frac{x_1^k}{k!}\left(\frac{\partial^2 f_k}{\partial x_2^2}+\frac{\partial^2 f_k}{\partial x_3^2}\right) \\ &= \frac{f_k(x_2,x_3)}{(k-2)!}x_1^{k-2}+\frac{x_1^k}{k!}\left(\frac{\partial^2 f_k}{\partial x_2^2}+\frac{\partial^2 f_k}{\partial x_3^2}\right) \end{aligned}

Now we can put them together naturally:

$\Delta f = \sum_{k=0}^{\ell-2}\frac{f_{k+2}}{k!}x_1^{k}+\sum_{k=0}^{\ell}\frac{x_1^k}{k!}\left(\frac{\partial^2 f_k}{\partial x_2^2}+\frac{\partial^2 f_k}{\partial x_3^2}\right)$

Let's try to explore the last term a little bit more. If $k=\ell-1$ or $\ell$, then $f_k$ is of order $0$ and $1$ and consequently the second order derivative is $0$. Therefore we write

$\Delta f = \sum_{k=0}^{\ell-2}\frac{x_1^k}{k!}\left[f_{k+2}+\left(\frac{\partial^2 f_k}{\partial x_2^2}+\frac{\partial^2 f_k}{\partial x_3^2}\right)\right]$

Therefore $\Delta{f}=0$ if and only if

$f_{k+2}+\left(\frac{\partial^2 f_k}{\partial x_2^2}+\frac{\partial^2 f_k}{\partial x_3^2}\right)=0.$

Therefore, once $f_0$ and $f_1$ is determined, all of $f_k$ are determined and so is $f$ itself. Therefore

$\dim \mathfrak{H}_\ell=\dim P_\ell^2+\dim P_{\ell-2}^2$

where $P_k^2$ is the space of homogeneous polynomials with two variables, hence is isomorphic to $\mathbb{C} \otimes_\mathbb{R}\operatorname{Sym}^k \mathbb{R}^2$, and we therefore have

$\dim P_\ell^2 = {\ell+2-1 \choose \ell}=\ell+1, \quad \dim P_{\ell-1}^2 = \ell.$

Hence

$\dim \mathfrak{H}_\ell=2\ell+1.$

$\square$

Recall that $\dim W_n=2n+1$. This should not be a coincidence, and we shall dive into it right now. To do this we immediately establish the connection between $\Delta$ and $SO(3)$.

Lemma 2. The action of the Laplacian on $C^\infty(\mathbb{R}^3;\mathbb{C})$ (which contains $P_\ell$ for all $\ell \ge 0$) commutes with the action of $SO(3)$, i.e. $\Delta$ is $SO(3)$-equivariant.

Proof. Really routine verification. $\square$

As a result, we have an very important result:

Theorem 1. The space $\mathfrak{H}_\ell$ is an $SO(3)$-invariant subspace of $P_\ell$.

## Characters of SO(3)

We start with a direct observation of matrices in $SO(3)$, that 'downgrades' the rotation to a plane:

Lemma 2. Every element in $SO(3)$ is conjugate to $R(t)$, where

$R(t)= \begin{pmatrix}1 & 0 & 0 \\ 0&\cos{t}&-\sin{t} \\ 0&\sin{t}&\cos{t} \end{pmatrix}.$

Proof. Pick any $A \in SO(3)$. First of all we show that $A$ has an eigenvalue $1$. Note

\begin{aligned} \det (I-A)&=\det(AA^T-A) \\ &=\det(A(A^T-I)) \\ &=\det(A)\det(A^T-I) \\ &=\det(A-I) \\ &=-\det(I-A) \end{aligned}

we therefore have $\det(I-A)=0$. Hence we can pick $v_1 \in \ker(I-A)$ with norm $1$. Pick $v_2 \in \mathbb{R}^3$ pedicular to $v_1$ with norm $1$ and $v_3=v_1 \times v_2$. Then $\{v_1,v_2,v_3\}$ is an orthonormal basis and $V=(v_1,v_2,v_3)$ is in $SO(3)$. $A$ is therefore conjugate to

$V^{-1}AV=R=\begin{pmatrix}1 & 0 & 0 \\ 0&a&b \\ 0&c&d \end{pmatrix} \in SO(3).$

In particular, $R \in SO(3)$ also implies

$\begin{cases} a^2+b^2=1 \\ b^2+d^2=1 \\ a^2+c^2=1 \\ b^2+d^2=1 \\ ad-bc=1 \end{cases}$

Solving this equation system we must have $a=d$, $b=-c$ so that we can assign $a=\cos{t}$ and $c=\sin{t}$, and the result follows. $\square$

Since characters are invariant under conjugation, the study of the character of $SO(3)$ is reduced to $T$, the subgroup generated by matrices of the form $R(t)$. But direct computation is a nightmare so we try our best to do it elegantly. To do this, we return to the irreducible representations of $SU(2)$ (there are only two variables, anyway). The canonical map $SU(2) \to SO(3)$ has a specific value:

$e(t)=\begin{pmatrix} \exp(it) & 0 \\ 0 & \exp(-it) \end{pmatrix} \mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos{2t} & -\sin{2t} \\ 0 & \sin{2t} & \cos{2t} \end{pmatrix}=R(2t).$

One can refer to this document for the map above. Our study of characters is now reduced to $SU(2)$, because $\chi_{W_n}(R(t))=\chi_{V_{2n}}(e(t/2))$, using the facts that character is invariant under isomorphism and that $V_{2n} \cong W_n$. We can compute that

$\chi_{V_{2n}}(e(t/2))=\sum_{k=0}^{2n}\exp\left(i(2n-2k)\frac{t}{2}\right)=\sum_{k=0}^{2n}\exp(i(n-k)t).$

Now we are ready for the irreducible representations of $SO(3)$.

# Determining All the Irreducible SO(3)-modules

Since we basically have $\dim \mathfrak{H}_\ell=\dim W_\ell$, it is natural to believe that $\mathfrak{H}_\ell \cong W_\ell$, in the sense of $SO(3)$-modules, and the following theorem answers this question affirmatively.

Theorem 2. The space $\mathfrak{H}_\ell$ is isomorphic to $W_\ell$. In other words, irreducible $SO(3)$-modules are determined by spherical harmonics.

Proof. We will use the fact every compact Lie group is completely reducible. (First of all, $SO(3)$ is compact as it is a closed subgroup of this document. On the other hand, the fact that every compact Lie group is completely reducible can be found in section 3 of this document).

Therefore we have

$\mathfrak{H}_\ell= \bigoplus_{\nu}W_{n_\nu}$

where each $W_{n_\nu}$ is an irreducible representation of $SO(3)$. Applying dimension on both sides yields

$2\ell+1 = \sum_{\nu}(2n_\nu+1).$

To prove that $\mathfrak{H}_\ell=W_{\ell}$, it suffices to show that $n_\nu \ge \ell$ for some $n_\nu$. On the other hand, applying characters on both side we see

$\chi_{\mathfrak{H}_\ell}(R(t))=\sum_{n_\nu}\sum_{k=-n_\nu}^{n_\nu}\exp(ikt).$

In other words, the character is a linear combination of $\exp(ikt)$ with $|k| \le \max_\nu n_\nu$ for all $k$. Each $\exp(ikt)$ appeared above is an eigenvalue of the action of $R(t)$. We have no idea about the distribution of $k$ but we don't have to because our job is done if we can show that the action of $R(t)$ has an eigenvalue $\exp(-i\ell t)$.

To do this, we can consider vector $f(x_1,x_2,x_3)=(x_2+ix_3)^\ell \in \mathfrak{H}_\ell$. This is because for this vector we have

\begin{aligned} (R(t)f_\ell)(t)&=\left(x_2\cos{t}+x_3\sin{t}+i(-x_2\sin{t}+x_3\cos{t})\right)^\ell \\ &=(e^{-it}x_2+ie^{-it}x_3)^\ell \\ &=e^{-i\ell t}(x_2+ix_3)^\ell \\ &=e^{-i\ell t}f_\ell(t). \end{aligned}

$\square$

# Ending Remarks

• We find the eigenvalue because it shows that $\exp(-i\ell t)$ appears in the summand of $\chi_{\mathfrak{H}_\ell}(R(t))$, hence $|-\ell|=\ell \le \max n_\nu$. Since $\{n_\nu\}$ is finite, the maximum can be attained, and therefore our argument on dimension is done.

• The representation of $U(2)$ can be deduced algebraically, for one only need to notice that $U(2) = (S^1 \times SU(2))/H$, where $H=\{(1,I),(-1,-I)\}$. One will also need an odd-even argument just like we did to $SO(3)$.

• Likewise, since $O(3)=SO(3) \times \mathbb{Z}/2\mathbb{Z}$, we can deduce the irreducible representations of $O(3)$ in a same fashion.

Irreducible Representations of SO(3) and the Laplacian

https://desvl.xyz/2022/06/16/so3-laplacian/

Desvl

2022-06-16

2022-06-18