Representation Theory IV 2 Representations of Lie groups and Lie algebras - generalities 4 SL₃

3 SL₂

In this section we discuss the finite-dimensional representation theory of the Lie algebra $\mathfrak{sl}_{2,\mathbb{C}}$ . We then use that to study the representation theory of $\operatorname{SL}_{2}(\mathbb{C})$ , $\mathfrak{sl}_{2,\mathbb{R}}$ , $\operatorname{SL}_{2}(\mathbb{R})$ , and $\mathrm{SU}(2)$ , which are all closely related.

3.1 Weights

We fix the following ’standard’ basis of $\mathfrak{sl}_{2,\mathbb{C}}$ :

	$\displaystyle H$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 1&0\\ \hidden@noalign{}\hfil\textstyle 0&-1\end{pmatrix},$
	$\displaystyle X$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&1\\ \hidden@noalign{}\hfil\textstyle 0&0\end{pmatrix},$
and
	$\displaystyle Y$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&0\\ \hidden@noalign{}\hfil\textstyle 1&0\end{pmatrix}.$

These satisfy the following commutation relations, which are fundamental (check them!):

	$\displaystyle[H,X]$	$\displaystyle=2X,$
	$\displaystyle[H,Y]$	$\displaystyle=-2Y,$
and
	$\displaystyle[X,Y]$	$\displaystyle=H.$

Our strategy is to decompose representations of $\mathfrak{sl}_{2,\mathbb{C}}$ into eigenspaces for the action of $H$ . The elements $X$ and $Y$ will then move vectors between these eigenspaces, and this will let us analyze the representation theory of $\mathfrak{sl}_{2,\mathbb{C}}$ .

Let $(\rho,V)$ be a finite-dimensional complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ . Since $\rho(H)$ is an endomorphism of a finite-dimensional complex vector space, it has at least one eigenvalue. For $\alpha\in\mathbb{C}$ , write

V_{\alpha}=\{v\in V:\rho(H)v=\alpha v\}

for the $\alpha$ -eigenspace of $\rho(H)$ .

Definition 3.1.

1.

An eigenvalue $\alpha$ of $\rho(H)$ is called a weight (more precisely, an $H$ -weight) for the representation $\rho$ .
2.

Each $V_{\alpha}$ is called a weight space for $\rho$ .
3.

The nonzero vectors in $V_{\alpha}$ are called weight vectors for $\rho$ .

We will see later (Remark 3.18) that $\rho(H)$ is in fact diagonalizable with integer eigenvalues, so that $V$ decomposes as a direct sum of its weight spaces. However, the classification of irreducible representations does not require this.

Example 3.2.

The set of weights of the zero representation is empty, while the trivial representation has a single weight, $0$ .

Example 3.3.

Let $V=\mathbb{C}^{2}$ be the standard representation. Write $e_{1},e_{-1}$ for the standard basis. Then $He_{1}=e_{1}$ and $He_{-1}=-e_{-1}$ . Thus the set of weights of $V$ is $\{\pm 1\}$ .

Example 3.4.

We consider the adjoint representation $\operatorname{ad}$ of $\mathfrak{g}=\mathfrak{sl}_{2,\mathbb{C}}$ on itself. By the commutation relations, we see directly that $\operatorname{ad}_{H}$ has eigenvalues $0$ ( $[H,H]=0$ ), $2$ ( $[H,X]=2X$ ), and $-2$ ( $[H,Y]=-2Y$ ), so the set of weights is

\{-2,0,2\}.

The non-zero weights $2$ and $-2$ are called the roots of $\mathfrak{sl}_{2,\mathbb{C}}$ and their weight spaces are the root spaces $\mathfrak{g}_{2}$ and $\mathfrak{g}_{-2}$ . The weight vectors are called root vectors.

Thus we have the root space decomposition

	$\displaystyle\mathfrak{sl}_{2,\mathbb{C}}$	$\displaystyle=\mathfrak{g}_{0}\oplus\mathfrak{g}_{2}\oplus\mathfrak{g}_{-2}$
		$\displaystyle=\left\langle H\right\rangle\oplus\left\langle X\right\rangle% \oplus\left\langle Y\right\rangle.$

Example 3.5.

We consider $\mathbb{C}^{2}\otimes\mathbb{C}^{2}$ where $\mathbb{C}^{2}$ is the standard representation. Then

H(e_{1}\otimes e_{1})=(He_{1})\otimes e_{1}+e_{1}\otimes He_{1}=2e_{1}\otimes e% _{1}

and similarly

H(e_{1}\otimes e_{-1})=H(e_{-1}\otimes e_{1})=0,H(e_{-1}\otimes e_{-1})=-2e_{-% 1}\otimes e_{-1}

so that the weights are $\{-2,0,0,2\}$ . Note that this is a multiset — a set with repeated elements — and we say that the weight 0 has ‘multiplicity two’ (in general, the multiplicity of a weight is the dimension of the weight space).

Example 3.6.

Take $V=\operatorname{Sym}^{k}(\mathbb{C}^{2})$ . A set of basis vectors is

\{e_{1}^{a}e_{-1}^{b}:a,b\geq 0,a+b=k\}.

We calculate

	$\displaystyle H(e_{1}^{a}e_{-1}^{b})$	$\displaystyle=a(He_{1})e_{1}^{a-1}e_{-1}^{b}+b(He_{-1})e_{1}^{a}e_{-1}^{b-1}$
		$\displaystyle=(a-b)e_{1}^{a}e_{-1}^{b}.$

Thus the weights are (writing $b=k-a$ ):

\{-k,2-k,4-k,\ldots,k-4,k-2,k\}.

We will soon see an explanation for this pattern.

3.1.1 Highest weights

The following is our first version of the fundamental weight calculation.

Lemma 3.7.

Let $(\rho,V)$ be a complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ . Let $\alpha$ be a weight of $V$ and let $v\in V_{\alpha}$ . Then

X(v)\in V_{\alpha+2}

and

Y(v)\in V_{\alpha-2}.

Thus we have three maps:

	$\displaystyle H$	$\displaystyle:V_{\alpha}\to V_{\alpha}$
	$\displaystyle X$	$\displaystyle:V_{\alpha}\to V_{\alpha+2}$
	$\displaystyle Y$	$\displaystyle:V_{\alpha}\to V_{\alpha-2}.$

Proof.

We have, for $v\in V_{\alpha}$ ,

	$\displaystyle\rho(H)\rho(X)v$	$\displaystyle=[\rho(H),\rho(X)]v+\rho(X)\rho(H)v$
		$\displaystyle=\rho([H,X])v+\rho(X)\rho(H)v$
		$\displaystyle=2\rho(X)v+\alpha\rho(X)v$
		$\displaystyle=(\alpha+2)\rho(X)v.$

So $\rho(X)v\in V_{\alpha+2}$ as required.

The claim about the action of $Y$ is proved similarly. ∎

Definition 3.8.

A vector $v\in V_{\alpha}$ is a highest weight vector if it is a weight vector and if

Xv=0.

In this case we call the weight of $v$ a highest weight.

Lemma 3.9.

Any nonzero finite-dimensional complex linear representation $V$ of $\mathfrak{sl}_{2,\mathbb{C}}$ has a highest weight vector.

Proof.

Since $V$ is finite-dimensional, $\rho(H)$ has at least one eigenvalue. Among all weights of $V$ , choose one, $\alpha$ , with maximal real part and let $v$ be a weight vector of weight $\alpha$ . Then $X v$ has weight $\alpha+2$ by the fundamental weight calculation. Since $\alpha+2$ has strictly greater real part than $\alpha$ , it cannot be a weight of $V$ , so $Xv=0$ . ∎

Example 3.10.

Let $V=\mathbb{C}^{2}\otimes\mathbb{C}^{2}$ . Then the highest weight vectors are $e_{1}\otimes e_{1}$ and $e_{1}\otimes e_{-1}-e_{-1}\otimes e_{1}$ .

These are easily checked to be highest weight vectors — the first is killed by $X$ since $Xe_{1}=0$ , the second becomes

e_{1}\otimes e_{1}-e_{1}\otimes e_{1}=0.

It is left to you to check that there are no further highest weight vectors.

3.2 Classification of representations of sl_2,C.

The key point here is that a highest weight vector must have non-negative integer weight $n$ , and it then generates an irreducible representation of dimension $n+1$ whose isomorphism class is determined by $n$ and which has a very natural basis of weight vectors.

Let $(\rho,V)$ be a complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ .

Lemma 3.11.

Suppose that $V$ has a highest weight vector $v$ of weight $n$ . Then the subspace $W$ spanned by the vectors

v,Y(v),Y^{2}(v)=Y(Y(v)),\dots

is an $\mathfrak{sl}_{2,\mathbb{C}}$ -invariant subspace of $V$ .

Moreover, $n\geq 0$ , the dimension of $W$ is $n+1$ , $v,Y(v),\ldots,Y^{n}(v)$ are a basis for $W$ , with $Y^{n+1}(v)=0$ .

Proof.

Let $W$ be the span of the $Y^{k}(v)$ . Since the $Y^{k}(v)$ are weight vectors, their span is $H$ -invariant. It is clearly also $Y$ -invariant. So we only need to check the invariance under the $X$ -action. I claim that for all $m\geq 1$ ,

XY^{m}(v)=m(n-m+1)Y^{m-1}(v).

(3.1)

The proof is by induction. The case $m=1$ is:

XY(v)=([X,Y]+YX)v=Hv+Y(Xv)=Hv=nv.

If the formula holds for $m$ , then

$\displaystyle XY^{m+1}(v)$	$\displaystyle=([X,Y]+YX)Y^{m}(v)$
	$\displaystyle=HY^{m}v+Y(XY^{m}v)$
	$\displaystyle=(n-2m)Y^{m}v+m(n-m+1)Y^{m}v$	(induction hypothesis)
	$\displaystyle=(m+1)(n-m)Y^{m}v$

as required.

If $n$ is not a nonnegative integer, then $m(n-m+1)\neq 0$ for all $m$ . So

Y^{m-1}v\neq 0\implies XY^{m}v\neq 0\implies Y^{m}v\neq 0,

whence $Y^{m}v\neq 0$ for all $m$ . As these are weight vectors with distinct weights, they are linearly independent and so span an infinite dimensional subspace.

If $Y^{i}v=0$ for some $0<i\leq n$ , then

0=XY^{i}v=i(n-i+1)Y^{i-1}v

and so $Y^{i-1}v=0$ . Repeating gives that $v=Y^{0}v=0$ , a contradiction.

Now,

XY^{n+1}(v)=(n+1)(n-n)v=0

and so $Y^{n+1}v$ is either zero or a highest weight vector of weight $-(n+2)<0$ . We have already seen that the second possibility cannot happen, so $Y^{n+1}v=0$ .

Thus $W$ is spanned by the (nonzero) weight vectors $v,Yv,\ldots,Y^{n}v$ with distinct weights, which are therefore linearly independent and so a basis for $W$ . ∎

Remark 3.12.

If we didn’t assume that $V$ was finite-dimensional, then the first part of the previous lemma would still be true.

Corollary 3.13.

In the situation of the previous lemma, $W$ is irreducible.

Proof.

Suppose that $W^{\prime}\subset W$ is a nonzero subrepresentation. Then it has a highest weight vector, which must be (proportional to) $Y^{i}v$ for some $0\leq i\leq n$ . But then

XY^{i}v=i(n-i+1)Y^{i-1}v\neq 0

if $i>0$ and so $i=0$ , meaning that $v\in W^{\prime}$ . But then $Y^{i}v\in W^{\prime}$ for all $i$ , so $W^{\prime}=W$ as required. ∎

The weights of $W$ are illustrated in Figure 3.2.

Theorem 3.14.

Suppose $V$ is an irreducible finite-dimensional complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ . Then:

1.

There is a unique (up to scalar) highest weight vector, with highest weight $n\geq 0$ .
2.

The weights of $V$ are $n,n-2,\ldots,2-n,-n$ .
3.

All weight spaces $V_{\alpha}$ are one-dimensional (we say that $V$ is ‘multiplicity free’).
4.

The dimension of $V$ is $n+1$ .

For every $n\geq 0$ , there is a unique irreducible complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ (up to isomorphism) with highest weight $n$ .

Proof.

Let $V$ be an irreducible finite-dimensional complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ . Let $v\in V$ be a highest weight vector. By Lemma 3.11 its weight is a nonnegative integer $n$ , and by Corollary 3.13 the vectors $v,Yv,\ldots,Y^{n}v$ span an irreducible subrepresentation of $V$ of dimension $n+1$ , which must therefore be the whole of $V$ . The claims all follow immediately.

Moreover, the actions of $X$ , $Y$ and $H$ on this basis are given by explicit matrices depending only on $n$ , so the isomorphism class of $V$ is determined by $n$ .

It remains only to show that a representation with highest weight $n$ exists for all $n\geq 0$ . Consider $\operatorname{Sym}^{n}(\mathbb{C}^{2})$ . The vector $e_{1}^{n}$ is a highest weight vector of weight $n$ , so we are done. In fact, in this case $Y^{a}e_{1}^{n}$ is proportional to $e_{1}^{n-a}e_{-1}^{a}$ , so the irreducible representation generated by $e_{1}^{n}$ is in fact the whole of $\operatorname{Sym}^{n}(\mathbb{C}^{2})$ . ∎

In fact, if $(\rho,V)$ is the irreducible representation with highest weight $n$ , highest weight vector $v$ , then the matrices of $H$ , $Y$ , and $X$ with respect to the basis

\{v,Y(v),\cdots,Y^{n}(v)\}

are respectively

	$\displaystyle\rho(H)$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle n&&&\\ \hidden@noalign{}\hfil\textstyle&n-2&&\\ \hidden@noalign{}\hfil\textstyle&&\ddots&\\ \hidden@noalign{}\hfil\textstyle&&&-n\end{pmatrix},$
	$\displaystyle\rho(Y)$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&&&\\ \hidden@noalign{}\hfil\textstyle 1&0&&\\ \hidden@noalign{}\hfil\textstyle&\ddots&\ddots&\\ \hidden@noalign{}\hfil\textstyle&&1&0\end{pmatrix},$
and
	$\displaystyle\rho(X)$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&n&&&&\\ \hidden@noalign{}\hfil\textstyle&0&2(n-1)&&&\\ \hidden@noalign{}\hfil\textstyle&&0&3(n-2)&&\\ \hidden@noalign{}\hfil\textstyle&&&\ddots&\ddots&\\ \hidden@noalign{}\hfil\textstyle&&&&\ddots&n\\ \hidden@noalign{}\hfil\textstyle&&&&&0\end{pmatrix}.$

Since $\operatorname{SL}_{2}(\mathbb{C})$ is simply connected, we have:

Proposition 3.15.

Every finite-dimensional complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ is the derivative of a unique representation of $\operatorname{SL}_{2}(\mathbb{C})$ .

Theorem 3.16.

Every finite-dimensional irreducible complex-linear representation of $\operatorname{SL}_{2}(\mathbb{C})$ or $\mathfrak{sl}_{2,\mathbb{C}}$ is isomorphic to $\operatorname{Sym}^{n}(\mathbb{C}^{2})$ , the symmetric power of the standard representation.

Proof.

We already proved this for $\mathfrak{sl}_{2,\mathbb{C}}$ . By Proposition 3.15, every representation of $\operatorname{SL}_{2}(\mathbb{C})$ arises from one of $\mathfrak{sl}_{2,\mathbb{C}}$ , and since $\operatorname{SL}_{2}(\mathbb{C})$ is connected the result follows. ∎

3.3 Decomposing representations

Theorem 3.17.

Every finite-dimensional complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ is completely reducible, that is, splits into a direct sum of irreducible representations.

Proof.

This is a special case of Theorem 3.27 below. ∎

Remark 3.18.

Complete reducibility shows that every finite-dimensional representation of $\mathfrak{sl}_{2,\mathbb{C}}$ is a direct sum of irreducibles. Each irreducible has integer weights and a basis of weight vectors. Thus these properties hold for any finite-dimensional representation: $\rho(H)$ is diagonalizable with integer eigenvalues and we get a weight space decomposition

V=\bigoplus_{n\in\mathbb{Z}}V_{n}.

(3.2)

One can also see this more directly: by Proposition 3.15, $\rho$ exponentiates to $\operatorname{SL}_{2}(\mathbb{C})$ . The subgroup $\operatorname{U}(1)\subset\operatorname{SL}_{2}(\mathbb{C})$ (embedded via $e^{it}\mapsto\begin{pmatrix}e^{it}&\\ &e^{-it}\end{pmatrix}$ ) acts completely reducibly by Maschke’s theorem (section 2.5), and its irreducible representations are $e^{it}\mapsto e^{int}$ for $n\in\mathbb{Z}$ (Theorem 2.18). Taking the derivative shows that $\rho(iH)$ is diagonalizable with eigenvalues in $i\mathbb{Z}$ and so $\rho(H)$ is diagonalizable with integer eigenvalues. The integrality of the weights thus reflects the fact that the characters of $\operatorname{U}(1)$ are indexed by integers.

It is easy to decompose a representation $V$ of $\mathfrak{sl}_{2,\mathbb{C}}$ into irreducibles by looking at the weights. Firstly, look at the maximal weight $k$ of $V$ . Then there must be a weight vector $v$ of weight $k$ , which is necessarily a highest weight vector, and so $V$ must contain a copy of $\operatorname{Sym}^{k}(\mathbb{C}^{2})$ — namely, the subspace $\left\langle v,Yv,\ldots,Y^{k}v\right\rangle$ . By complete reducibility we have

V\cong\operatorname{Sym}^{k}(\mathbb{C}^{2})\oplus W.

The weights of $W$ are then obtained by removing the weights of $\operatorname{Sym}^{k}(\mathbb{C}^{2})$ from the weights of $V$ , and we repeat the process.

In particular, this shows that a finite-dimensional representation of $\mathfrak{sl}_{2,\mathbb{C}}$ is determined, up to isomorphism, by its multiset of weights.

Proposition 3.19.

If $V$ , $W$ are representations of $\mathfrak{sl}_{2,\mathbb{C}}$ then:

•

$\{$ weights of $V\otimes W\}=\{$ weights of $V\}+\{$ weights of $W\}$ .
•

$\{$ weights of $\operatorname{Sym}^{k}(V)\}=\{$ sums of unordered $k$ -tuples of weights of $V\}$ .
•

$\{$ weights of $\Lambda^{k}(V)\}=\{$ sums of unordered ‘distinct’ $k$ -tuples of weights of $V\}$ .

Proof.

If $v_{1},\ldots,v_{m}$ is a basis of weight vectors of $V$ with $Hv_{i}=\alpha_{i}v_{i}$ , and $w_{1},\ldots,w_{n}$ is a basis of weight vectors of $W$ with $Hw_{i}=\beta_{i}v_{i}$ , then

\{v_{i}\otimes w_{j}:1\leq i\leq m,1\leq j\leq n\}

is a basis of weight vectors of $V\otimes W$ and

H(v_{i}\otimes w_{j})=(Hv_{i})\otimes w_{j}+v_{i}\otimes HW_{j}=(\alpha_{i}+% \beta_{j})(v_{i}\otimes w_{j}).

This shows part (1), and the other parts are very similar.

Example 3.20.

We should illustrate what is meant by ‘distinct’: it is ‘distinct’ as elements of the multiset. Suppose that the weights of $V$ are $\{2,0,0,-2\}$ . Then to obtain the weights of $\Lambda^{2}(V)$ we add together unordered, distinct, pairs of these in every possible way, getting:

\{2+0,2+0,2+-2,0+0,0+-2,0+-2\}=\{2,2,0,0,-2,-2\}.

Example 3.21.

Let $\mathbb{C}^{2}$ be the standard representation of $\operatorname{SL}_{2}(\mathbb{C})$ with weight basis $e_{1}$ , $e_{-1}$ . Consider $V=\operatorname{Sym}^{2}(\mathbb{C}^{2})\otimes\operatorname{Sym}^{2}(\mathbb{% C}^{2})$ .

The weights of $\operatorname{Sym}^{2}(\mathbb{C}^{2})$ are $\{-2,0,2\}$ and so the weights of $V$ are

\{-2,0,2\}+\{-2,0,2\}=\{-4,-2,-2,0,0,0,2,2,4\}.

This is the same as the set of weights of

\operatorname{Sym}^{4}(\mathbb{C}^{2})\oplus\operatorname{Sym}^{2}(\mathbb{C}^% {2})\oplus\mathbb{C}

and so this is the required decomposition into irreducibles.

Figure 4: Decomposing $\operatorname{Sym}^{2}(\mathbb{C}^{2})\otimes\operatorname{Sym}^{2}(\mathbb{C}% ^{2})$ .

We can go further, and decompose $V$ into irreducible subrepresentations. This means finding irreducible subrepresentations of $V$ such that $V$ is their direct sum. To start with we write down the weight vectors:

Let $v_{2}=e_{1}^{2}$ , $v_{0}=e_{1}e_{-1}$ and $v_{-2}=e_{-1}^{2}$ be weight vectors in $\operatorname{Sym}^{2}(\mathbb{C}^{2})$ corresponding to the weights $2$ , $0$ and $-2$ . Then the weights of $V$ are

•

$4$ , multiplicity one, weight vector: $v_{2}\otimes v_{2}$ .
•

$2$ , multiplicity two, weight space: $\left\langle v_{2}\otimes v_{0},v_{0}\otimes v_{2}\right\rangle$ .
•

$0$ , multiplicity three, weight space: $\left\langle v_{2}\otimes v_{-2},v_{0}\otimes v_{0},v_{-2}\otimes v_{2}\right\rangle$ .
•

$-2$ , multiplicity two, weight space: $\left\langle v_{0}\otimes v_{-2},v_{-2}\otimes v_{0}\right\rangle$ .
•

$-4$ , multiplicity one, weight vector: $\left\langle v_{-2}\otimes v_{-2}\right\rangle$ .

The copy of $\operatorname{Sym}^{4}(\mathbb{C}^{2})$ in $V$ has highest weight vector $v_{2}\otimes v_{2}$ . We can find a basis by repeatedly hitting this with $Y$ (writing $\propto$ for ‘equal up to a nonzero scalar’):

	$\displaystyle Y(v_{2}\otimes v_{2})$	$\displaystyle=2(v_{2}\otimes v_{0}+v_{0}\otimes v_{2})\propto v_{2}\otimes v_{% 0}+v_{0}\otimes v_{2}$
	$\displaystyle Y^{2}(v_{2}\otimes v_{2})$	$\displaystyle\propto v_{2}\otimes v_{-2}+4v_{0}\otimes v_{0}+v_{-2}\otimes v_{2}$
	$\displaystyle Y^{3}(v_{2}\otimes v_{2})$	$\displaystyle\propto 6(v_{0}\otimes v_{-2}+v_{-2}\otimes v_{0})\propto v_{0}% \otimes v_{-2}+v_{-2}\otimes v_{0}$
	$\displaystyle Y^{4}(v_{2}\otimes v_{2})$	$\displaystyle\propto v_{-2}\otimes v_{-2}.$

These vectors are a basis for the copy of $\operatorname{Sym}^{4}(\mathbb{C}^{2})$ in $V$ .

Next, we find the copy of $\operatorname{Sym}^{2}(\mathbb{C}^{2})$ in $V$ . We start by looking for a highest weight vector of weight 2:

v_{2}\otimes v_{0}-v_{0}\otimes v_{2}

does the trick. Hitting this with $Y$ gives $v_{2}\otimes v_{-2}-v_{-2}\otimes v_{2}$ , and doing so again gives $v_{-2}\otimes v_{0}-v_{0}\otimes v_{-2}$ (up to scalar). These vectors are a basis for the copy of $\operatorname{Sym}^{2}(\mathbb{C}^{2})$ in $V$ .

Finally, we find the trivial representation $\mathbb{C}$ in $V$ . We need only find a weight vector of weight 0 which is killed by $X$ , and

v_{2}\otimes v_{-2}-2v_{0}\otimes v_{0}+v_{-2}\otimes v_{2}

does the job: this vector spans a copy of the trivial representation.

3.4 Real forms and complete reducibility

We use our understanding of the representation theory of $\mathfrak{sl}_{2,\mathbb{C}}$ to understand the representation theory of $\operatorname{SL}_{2}(\mathbb{R})$ and $\mathrm{SU}(2)$ , and prove complete reducibility.

Definition 3.22.

A real form of a complex Lie algebra $\mathfrak{g}$ is a real Lie algebra $\mathfrak{h}\subset\mathfrak{g}$ such that every element $Z$ of $\mathfrak{g}$ can be written uniquely as $X+iY$ for $X,Y\in\mathfrak{h}$ .

For dimension reasons, necessary and sufficient conditions are that $\mathfrak{h}\cap i\mathfrak{h}=0$ and $\dim_{\mathbb{R}}\mathfrak{h}=\dim_{\mathbb{C}}\mathfrak{g}$ .

Example 3.23.

The Lie algebra $\mathfrak{sl}_{n,\mathbb{R}}$ is a real form of $\mathfrak{sl}_{n,\mathbb{C}}$ .

Example 3.24.

The Lie algebra $\mathfrak{su}_{n}$ is a real form of $\mathfrak{sl}_{n,\mathbb{C}}$ . Indeed, $\dim_{\mathbb{R}}\mathfrak{su}_{n}=n^{2}-1=\dim_{\mathbb{C}}\mathfrak{sl}_{n,% \mathbb{C}}$ and if $X,iX\in\mathfrak{su}_{n}$ then

iX=-(iX)^{\dagger}=iX^{\dagger}=-iX

so $X=0$ .

Explicitly, we may write $A\in\mathfrak{sl}_{n,\mathbb{C}}$ as $X+iY$ with

X=\frac{1}{2}(A-A^{\dagger})

and

Y=\frac{-i}{2}(A+A^{\dagger})

in $\mathfrak{su}_{n}$ .

Proposition 3.25.

Let $\mathfrak{h}$ be a real form of $\mathfrak{g}$ .

There is a one-to-one correspondence between representations of $\mathfrak{h}$ and complex-linear representations of $\mathfrak{g}$ under which irreducible representations correspond to irreducible representations.

Proof.

Given a $\mathbb{C}$ -linear representation of $\mathfrak{g}$ , it is a representation of $\mathfrak{h}$ by restriction. Conversely, if $(\rho,V)$ is a representation of $\mathfrak{h}$ , then it extends to a unique $\mathbb{C}$ -linear representation of $\mathfrak{g}$ given by the formula (forced by $\mathbb{C}$ -linearity)

\rho(X+iY)v=\rho(X)v+i\rho(Y)v.

It is easy to see that this preserves the Lie bracket. The proposition follows (the final statement is left as an exercise). ∎

As a corollary we immediately obtain

Theorem 3.26.

The representation theories of $\mathfrak{sl}_{n,\mathbb{R}}$ and $\mathfrak{su}_{n}$ are ‘the same’ as the complex-linear representation theory of $\mathfrak{sl}_{n,\mathbb{C}}$ . All finite-dimensional irreducible representations of $\mathfrak{sl}_{2,R}$ , $\operatorname{SL}_{2}(\mathbb{R})$ , $\mathfrak{su}_{2}$ , or $\mathrm{SU}(2)$ are of the form $\operatorname{Sym}^{n}(\mathbb{C}^{2})$ .

Proof.

The claims about Lie algebras follow from the above discussion. Every (finite-dimensional) irreducible representation of $\mathfrak{sl}_{2,\mathbb{R}}$ is of the form $\operatorname{Sym}^{n}(\mathbb{C}^{2})$ , and these clearly exponentiate to representations of $\operatorname{SL}_{2}(\mathbb{R})$ , despite this not being a simply connected group! Similarly for $\mathrm{SU}(2)$ (which is simply connected). Since $\operatorname{SL}_{2}(\mathbb{R})$ and $\mathrm{SU}(2)$ are connected, every representation of them is determined by its derivative, so we have a complete list of the irreducible representations. ∎

Theorem 3.27.

Every finite-dimensional complex-linear representation of $\mathfrak{sl}_{n,\mathbb{C}}$ is completely reducible.

Proof.

Let $V$ be a finite-dimensional complex-linear representation of $\mathfrak{sl}_{n,\mathbb{C}}$ and let $W\subset V$ be a subrepresentation. Then $W$ is an $\mathfrak{su}_{n}$ -subrepresentation. As $\mathrm{SU}(n)$ is simply-connected, $V$ and $W$ exponentiate to representations of $\mathrm{SU}(n)$ . Since $\mathrm{SU}(n)$ is compact, by Maschke’s theorem there is a complementary $\mathrm{SU}(n)$ -subrepresentation $W^{\prime}$ with

V=W\oplus W^{\prime}.

Then $W^{\prime}$ is a $\mathfrak{su}_{n}$ -subrepresentation, and so (since $\mathfrak{su}_{n}$ is a real form of $\mathfrak{sl}_{n,\mathbb{C}}$ ) a $\mathbb{C}$ -linear $\mathfrak{sl}_{n,\mathbb{C}}$ -subrepresentation. Complete reducibility follows. ∎

The argument in this proof is called Weyl’s unitary trick.

3.5 SO(3)

3.5.1 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 3.28.

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 $\mathrm{SU}(2)$ from section 2.4.

There is a surjective homomorphism $\pi:\mathrm{SU}(2)\to\mathrm{SO}(3)$ , given by the adjoint action of $\mathrm{SU}(2)$ on its Lie algebra endowed with a certain symmetric positive definite bilinear form. The kernel of $\pi$ is $\{\pm I\}$ .

Theorem 3.29.

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 I\}\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. ∎

There is an orthonormal basis $\mathcal{I},\mathcal{J},\mathcal{K}$ 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. We have computed in 2.4 that, under the (adjoint) map $D\pi$ , with respect to this basis,

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

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

With this calculation in hand, we can consider the weights of the representations in 3.29. 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 3.30.

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 $V^{(1)}$ .

3.5.2 Harmonic functions

Remark 3.31.

We went through this very quickly in the lecture; I have marked as nonexaminable the proofs that we skipped.

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 $\mathrm{SO}(3)$ and hence also of $\mathfrak{so}_{3}$ .

Lemma 3.32.

The elements $J_{x},J_{y}$ , and $J_{z}$ of $\mathfrak{so}_{3}$ 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.

(nonexaminable) 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 $x_{1},x_{2},x_{3}$ ). ∎

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 3.33.

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 3.34.

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

Proof.

(Nonexaminable.) 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 $\dim(\mathcal{P}^{\ell})>\dim(\mathcal{P}^{\ell-2})$ , nonzero harmonic polynomials must exist for all $\ell$ . We write $\mathcal{H}^{\ell}\subset\mathcal{P}^{\ell}$ for the space of harmonic polynomials.

Lemma 3.35.

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

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

Proof.

Left as a nonexaminable exercise. You may find the following useful:

Lemma 3.36.

(Euler’s formula) If $f\in\mathcal{P}^{\ell}$ , then

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

∎

It follows that $r^{2}\Delta$ preserves every 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 3.37.

nonexaminable Suppose that $V\subset\mathcal{P}^{\ell}$ is an irreducible subrepresentation 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 3.38.

For every $\ell\geq 2$ ,

\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 $\mathcal{P}^{\ell}$ has the following decomposition into irreducible subrepresentations:

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

Proof.

(Nonexaminable.) 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 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. ∎

Thus every homogeneous 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 3.39.

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 analogous to the Fourier decomposition of a function on the unit circle.

Example 3.40.

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}.

3 SL2

3.1 Weights

Definition 3.1.

Example 3.2.

Example 3.3.

Example 3.4.

Example 3.5.

Example 3.6.

3.1.1 Highest weights

Lemma 3.7.

Proof.

Definition 3.8.

Lemma 3.9.

Proof.

Example 3.10.

3.2 Classification of representations of sl2,C.

Lemma 3.11.

Proof.

Remark 3.12.

Corollary 3.13.

Proof.

Theorem 3.14.

Proof.

Proposition 3.15.

Theorem 3.16.

Proof.

3.3 Decomposing representations

Theorem 3.17.

Proof.

Remark 3.18.

Proposition 3.19.

Proof.

Example 3.20.

Example 3.21.

3.4 Real forms and complete reducibility

Definition 3.22.

Example 3.23.

Example 3.24.

Proposition 3.25.

Proof.

Theorem 3.26.

Proof.

Theorem 3.27.

Proof.

3.5 SO(3)

3.5.1 Classification of irreducible representations

Theorem 3.28.

Theorem 3.29.

Proof.

Example 3.30.

3.5.2 Harmonic functions

Remark 3.31.

Lemma 3.32.

Proof.

Lemma 3.33.

Proof.

Lemma 3.34.

Proof.

Lemma 3.35.

Proof.

Lemma 3.36.

Lemma 3.37.

Proof.

Theorem 3.38.

Proof.

Lemma 3.39.

Proof.

Example 3.40.

3 SL₂

3.2 Classification of representations of sl_2,C.