6 SL₂ 6.4 Real forms and complete decomposability 7 SL₃

6.5 SO(3) (non-examinable)

We skipped this section, due to the strike, so the material is non-examinable.

6.5.1 Classification of irreducible representations

We have already (see the first problem set this term) seen that $\mathfrak{su}_{2}$ and $\mathfrak{so}_{3}$ are isomorphic. We therefore have:

Theorem 6.31.

There is an irreducible two-dimensional representation $V$ of $\mathfrak{so}_{3}$ (coming from $\mathfrak{so}_{3}\xrightarrow{\sim}\mathfrak{su}_{2}\subset\mathfrak{gl}_{2,% \mathbb{C}}$ ) such that the irreducible complex representations of $\mathfrak{so}_{3}$ are exactly

\operatorname{Sym}^{n}(V)

for $n\geq 0$ .

We want to know which of these representations exponentiate to an irreducible representation of $\mathrm{SO}(3)$ . For this, we revisit the connection with $\mathrm{SU}(2)$ .

Let $\left\langle\cdot,\cdot\right\rangle$ be the bilinear form on $\mathfrak{su}_{2}$ defined by

\left\langle X,Y\right\rangle=-\operatorname{tr}(XY).

Lemma 6.32.

The form $\left\langle\cdot,\cdot\right\rangle$ is a positive definite bilinear form preserved by the adjoint action of $\mathrm{SU}(2)$ .

Lemma 6.33.

The bilinearity is clear. To see that it is positive definite, if $X\in\mathfrak{su}_{2}$ then

\left\langle X,X\right\rangle=-\operatorname{tr}(X^{2})=\operatorname{tr}(XX^{% \dagger}).

Direct calculation shows $\operatorname{tr}(XX^{\dagger})$ is the sum of the squares of the entries of $X$ , which is strictly positive for nonzero $X$ .

We have, for $g\in\mathrm{SU}(2)$ ,

\left\langle\operatorname{Ad}_{g}X,\operatorname{Ad}_{g}Y\right\rangle=-% \operatorname{tr}(gXg^{-1}gYg^{-1}=-\operatorname{tr}(gXYg^{-1})=-% \operatorname{tr}(XY)

which shows that this inner product is $\mathrm{SU}(2)$ -invariant.

Corollary 6.34.

There is an isomorphism

\mathrm{SU}(2)/\{\pm I\}\to\mathrm{SO}(3).

Proof.

Choosing an orthonormal basis for $\mathfrak{su}_{2}$ with respect to $\left\langle\cdot,\cdot\right\rangle$ identifies the set of linear maps $\mathfrak{su}_{2}\to\mathfrak{su}_{2}$ preserving the inner product with $\mathrm{SO}(3)$ . We therefore have a homomorphism

\mathrm{SU}(2)\to\mathrm{SO}(3)

whose kernel is $\{g\in\mathrm{SU}(2):\operatorname{Ad}_{g}=\operatorname{Ad}\}=\{\pm I\}$ . The derivative of this homomorphism is injective, otherwise there would be a one-parameter subgroup in the kernel of the group homomorphism, and since the Lie algebras have the same dimension we get an isomorphism of Lie algebras. The group homomorphism is therefore surjective, since exponentials of elements of $\mathfrak{so}_{3}$ generate the group $\mathrm{SO}(3)$ . ∎

Remark 6.35.

Let $\mathcal{I},\mathcal{J},\mathcal{K}$ be the elements of $\mathfrak{su}_{2}$ given by

\frac{1}{2}\begin{pmatrix}0&1\\ -1&0\end{pmatrix},\frac{1}{2}\begin{pmatrix}0&i\\ i&0\end{pmatrix},\frac{1}{2}\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}

respectively. Up to scalar, these are an orthonormal basis for $\mathfrak{su}_{2}$ . Since the derivative of the above homomorphism $\mathrm{SU}(2)\to\mathrm{SO}(3)$ is the adjoint map, computed with respect to this basis, we see that the induced isomorphism $\mathfrak{su}_{2}\to\mathfrak{so}_{3}$ takes:

	$\displaystyle\mathcal{I}\mapsto J_{x}$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&0&0\\ \hidden@noalign{}\hfil\textstyle 0&0&-1\\ \hidden@noalign{}\hfil\textstyle 0&1&0\end{pmatrix}$
	$\displaystyle\mathcal{J}\mapsto J_{y}$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&0&1\\ \hidden@noalign{}\hfil\textstyle 0&0&0\\ \hidden@noalign{}\hfil\textstyle-1&0&0\end{pmatrix}$
	$\displaystyle\mathcal{K}\mapsto J_{z}$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&-1&0\\ \hidden@noalign{}\hfil\textstyle 1&0&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0\end{pmatrix}.$

It will be useful to know where this isomorphism takes the elements $H, X, Y$ of $\mathfrak{sl}_{2,\mathbb{C}}=\mathfrak{su}_{2,\mathbb{C}}$ . For example, as $H=-2i\mathcal{K}$ , we see that it goes to $-2iJ_{z}$ . Or, for the lowering operator $Y$ , we have

Y=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}=-(\mathcal{I}+i\mathcal{J})\mapsto-(J_{x}+iJ_{y})

and similarly $X\mapsto J_{x}-iJ_{y}$ .

Corollary 6.36.

For each $\ell\geq 0$ , there is a unique irreducible representation $V^{(\ell)}$ of $\mathrm{SO}(3)$ of dimension $2\ell+1$ .

The derivative of this representation is isomorphic to the representation $\operatorname{Sym}^{2\ell}(V)$ of $\mathfrak{so}_{3}$ .

This gives the complete list of irreducible representations of $\mathrm{SO}(3)$ up to isomorphism.

Proof.

We simply have to work out which representations $\operatorname{Sym}^{k}(V)$ of $\mathfrak{so}_{3}$ exponentiate to a representation of $\mathrm{SO}(3)$ . Since we have

\mathrm{SU}(2)/\pm{1}\xrightarrow{\sim}\mathrm{SO}(3)

and each $\operatorname{Sym}^{k}(V)$ exponentiates to a unique representation of $\mathrm{SU}(2)$ — which we also call $\operatorname{Sym}^{k}(V)$ — this is equivalent to asking for which $k$ the centre $\{\pm I\}$ of $\mathrm{SU}(2)$ acts trivially on $\operatorname{Sym}^{k}(V)$ . But we see that $-I$ acts as $(-1)^{k}$ , so the answer is: for even $k$ only.

Thus $\operatorname{Sym}^{k}(V)$ exponentiates to a representation of $\mathrm{SO}(3)$ if, and only if, $k=2\ell$ is even, and we obtain the result. ∎

We can consider the weights of these representations. Under the isomorphism

\mathfrak{su}_{2,\mathbb{C}}=\mathfrak{sl}_{2,\mathbb{C}}\to\mathfrak{so}_{3,% \mathbb{C}}

the element $H=-2i\mathcal{K}$ maps to $-2iJ_{z}$ . Since the $H$ -weights of $V^{(\ell)}$ are $-2\ell,-2(\ell-1),\ldots,2(\ell-1),2\ell$ , we must divide these by $-2i$ to find the weights of $J_{z}$ acting on $V^{(\ell)}$ :

-i\ell,-i(\ell-1),\ldots,i(\ell-1),i\ell.

Example 6.37.

Consider the standard three-dimensional representation of $\mathrm{SO}(3)$ on $\mathbb{C}^{3}$ . The weights of $J_{z}$ are simply its eigenvalues as a $3\times 3$ matrix, which are $-i,0,i$ . We see that this representation is isomorphic to $V^{(1)}$ .

6.5.2 Harmonic functions

We can use our understanding of the representation theory of $\mathrm{SO}(3)$ to shed light on the classical theory of spherical harmonics.

We let $\mathcal{P}^{\ell}$ be the subspace of $\mathbb{C}[x,y,z]$ consisting of homogeneous polynomials of degree $\ell$ . This has an action of $\mathrm{SO}(3)$ given by

(gf)(\mathbf{x})=f(g^{T}\mathbf{x})

where $\mathbf{x}=\begin{pmatrix}x\\ y\\ z\end{pmatrix}$ is a vector in $\mathbb{C}^{3}$ . We therefore get a representation of $\mathrm{SO}(3)$ .

Lemma 6.38.

The elements $J_{x},J_{y}$ , and $J_{z}$ of $\mathrm{SO}(3)$ act on $\mathcal{P}^{\ell}$ according to the following formulae:

	$\displaystyle J_{x}$	$\displaystyle=z\frac{\partial}{\partial y}-y\frac{\partial}{\partial z}$
	$\displaystyle J_{y}$	$\displaystyle=x\frac{\partial}{\partial z}-z\frac{\partial}{\partial x}$
	$\displaystyle J_{z}$	$\displaystyle=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}.$

Proof.

Exercise. In fact, prove that the action of $A=(a_{ij})\in\mathfrak{so}_{3}$ is given by

\sum_{i,j}a_{ij}x_{i}\frac{\partial}{\partial x_{j}}

by considering

\frac{d}{dt}f(\exp(tA^{T})\mathbf{x})=\frac{d}{dt}f((I+tA^{T})\mathbf{x})

at $t=0$ (where we rewrite $x, y, z$ as $x_{1},x_{2},x_{3}$ ). ∎

It will be useful in what follows to recall

Lemma 6.39.

(Euler’s formula) If $f$ is a homogeneous polynomial of degree $\ell$ , then

x\frac{\partial}{\partial x}{f}+y\frac{\partial}{\partial y}{f}+z\frac{% \partial}{\partial z}{f}=\ell f.

The representation $\mathcal{P}^{\ell}$ is not irreducible. Let $r^{2}\in\mathbb{C}[x,y,z]$ be the polynomial

r^{2}=x^{2}+y^{2}+z^{2}.

Note that $r^{2}$ is clearly invariant under the action of $\mathrm{SO}(3)$ .

Lemma 6.40.

The map $\mathcal{P}^{\ell}\to\mathcal{P}^{\ell+2}$ defined by

f\mapsto r^{2}f

is an injective homomorphism of $\mathrm{SO}(3)$ -representations.

Proof.

We have, for $g\in\mathrm{SO}(3)$ ,

g(r^{2}f)=g(r^{2})g(f)=r^{2}g(f)

as required. ∎

Next, we consider the Laplace operator:

\Delta f=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2% }}+\frac{\partial^{2}}{\partial z^{2}}.

This is map from $\mathcal{P}^{\ell}\to\mathcal{P}^{\ell-2}$ .

Lemma 6.41.

The map $\Delta$ is a map of $\mathrm{SO}(3)$ representations.

Proof.

We must show that, for $g=(g_{ij})\in\mathrm{SO}(3)$ ,

(\Delta f)(g^{T}\mathbf{x})=\Delta(gf)(\mathbf{x})).

We have

\frac{\partial}{\partial x_{i}}(gf)(x)=\sum_{j}g_{ij}\frac{\partial f}{% \partial x_{j}}(g^{T}x)

and so

\frac{\partial}{\partial x_{i}}\frac{\partial}{\partial x_{i}}(gf)(x)=\sum_{j,% k}g_{ij}g_{ik}\frac{\partial}{\partial x_{k}}\frac{\partial}{\partial x_{j}}(g% ^{T}x).

We sum over $i$ , for fixed $j$ and $k$ :

\sum_{i}g_{ij}g_{ik}=\delta_{jk}

since $g$ is orthogonal. We therefore obtain

\sum_{i}\frac{\partial^{2}}{\partial x_{i}^{2}}(gf)(x)=\sum_{j}\frac{\partial^% {2}}{\partial x_{j}^{2}}(f)(g^{T}x),

and therefore

\Delta(gf)=g\Delta(f).

∎

An element $f\in\mathcal{P}^{\ell}$ is harmonic if $\Delta f=0$ . Since $\dim(\mathcal{P}^{\ell})>\dim(\mathcal{P}^{\ell-2})$ , harmonic polynomials must exist. We write $\mathcal{H}^{\ell}\subset\mathcal{P}^{\ell}$ for the space of harmonic polynomials.

Lemma 6.42.

On $\mathcal{P}^{\ell}$ , we have

r^{2}\Delta=J_{x}^{2}+J_{y}^{2}+J_{z}^{2}+\ell^{2}+\ell.

Proof.

Left as an exercise. ∎

It follows that $r^{2}\Delta$ preserves each irreducible subrepresentation of $\mathcal{P}^{\ell}$ (since $J_{x},J_{y},J_{z}$ do). Furthermore, by Schur’s lemma it must act on each irreducible subrepresentation as a scalar. We determine that scalar.

Lemma 6.43.

Suppose that $V\subset\mathcal{P}^{\ell}$ is an irreducible subrepresentattion with highest weight $i k$ . Then

(r^{2}\Delta)(f)=(\ell^{2}+\ell-k^{2}-k)f=(\ell-k)(\ell+k+1)f.

Proof.

Since $r^{2}\Delta$ is a $\mathfrak{so}_{3}$ -homomorphism and $V$ is irreducible, by Schur’s lemma it acts as a scalar on $V$ . It therefore suffices to compute the action on a highest weight vector $v\in V$ . So

J_{z}v=ikv,(J_{x}-iJ_{y})v=0.

It follows that $J_{z}^{2}v=-k^{2}v$ and, as

	$\displaystyle J_{x}^{2}+J_{y}^{2}$	$\displaystyle=(J_{x}+iJ_{y})(J_{x}-iJ_{y})+i[J_{x},J_{y}]$
		$\displaystyle=(J_{x}+iJ_{y})(J_{x}-iJ_{y})+iJ_{z},$

we have $(J_{x}^{2}+J_{y}^{2})v=0v-kv$ .

Applying the previous lemma gives the result. ∎

Theorem 6.44.

For every $\ell\geq 0$ ,

\mathcal{P}^{\ell}=\mathcal{H}^{\ell}\oplus r^{2}\mathcal{P}^{\ell-2}.

The space $\mathcal{H}^{\ell}$ is the irreducible highest weight representation of $\mathrm{SO}(3)$ of dimension $2\ell+1$ , and the space $\mathcal{P}^{\ell}$ , as an $\mathrm{SO}(3)$ -representation, the direct sum of representations of weights $i\ell,i(\ell-2),\ldots$ , each occuring with multiplicity one.

Proof.

We use induction on $\ell$ . The case $\ell=0$ is clear (we just have the trivial representation). Suppose true for $\ell-1$ with $\ell\geq 1$ .

By the previous lemma, the space $\mathcal{H}^{\ell}$ is the sum of all the copies inside $\mathcal{P}^{\ell}$ of the irreducible representation with with highest weight $i\ell$ . Since $\mathcal{P}^{\ell-2}$ does not contain this irreducible representation, by the inductive hypothesis, we have

\mathcal{H}^{\ell}\cap r^{2}\mathcal{P}^{\ell-2}=\{0\}.

Since, $\mathcal{H}^{\ell}\neq 0$ as already discussed, its dimension is a positive multiple of $2\ell+1$ . However, its dimension is at most

\dim\mathcal{P}^{\ell}-\dim\mathcal{P}^{\ell-2}=\binom{\ell+2}{2}-\binom{\ell}% {2}=2\ell+1.

It follows that $\mathcal{H}^{\ell}$ is irreducible, and that we have

\mathcal{H}^{\ell}\oplus r^{2}\mathcal{P}^{\ell-2}=\mathcal{P}^{\ell}.

The statement about the decomposition into irreducibles follows. ∎

The proof of this theorem shows that

\mathcal{P}^{\ell}=\mathcal{H}^{\ell}\oplus r^{2}\mathcal{H}^{\ell-2}\oplus r^% {4}\mathcal{P}^{\ell-4}\ldots

so that every polynomial has a unique decomposition as a sum of harmonic polynomials multiplied by powers of $r^{2}$ .

We can go further and give nice bases for the $\mathcal{H}^{\ell}$ by taking weight vectors for $J_{z}$ . First, we have

Lemma 6.45.

The function $(x-iy)^{\ell}\in\mathcal{P}^{\ell}$ is a highest weight vector of weight $i\ell$ .

Proof.

Exercise! ∎

We then obtain a weight basis by repeatedly applying the lowering operator

J_{x}+iJ_{y}=(ix-y)\frac{\partial}{\partial z}{}+z(\frac{\partial}{\partial y}% {}-i\frac{\partial}{\partial x}{}).

The functions thus obtained are known as ’spherical harmonics’ (at least, up to normalization), and give a particularly nice basis for the space of functions on the sphere $S^{2}\subset\mathbb{R}^{3}$ . The decomposition of a function into spherical harmonics is analagous to the Fourier decomposition of a function on the unit circle.

Example 6.46.

If $\ell=1$ , then $\mathcal{P}^{\ell}=\mathcal{H}^{\ell}$ , and the weight vectors are

x+iy,z,x-iy.

If $\ell=2$ , a basis of $\mathcal{H}^{\ell}$ made up of weight vectors is

(x-iy)^{2},z(x-iy),x^{2}+y^{2}-2z^{2},z(x+iy),(x+iy)^{2}.