3.12. Lecture 12

3.12.1. Weights

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

	$\displaystyle H$	$\displaystyle=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix},$
	$\displaystyle X$	$\displaystyle=\begin{pmatrix}0&1\\ 0&0\end{pmatrix},$
and
	$\displaystyle Y$	$\displaystyle=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}.$

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

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

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

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

Proposition 3.12.1.

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

Note that we have not proved this. However, we will use the result freely in what follows. It is possible to give purely algebraic proofs of all the results for which we use the previous proposition, but it is more complicated.

Proposition 3.12.2.

Let $(\rho,V)$ be a finite-dimensional complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ . Then $\rho(H)$ is diagonalisable with integer eigenvalues.

Proof.

By the previous proposition $\rho$ is the derivative of a representation $\tilde{\rho}$ of $\operatorname{SL}_{2}(\mathbb{C})$ . We can identify $\operatorname{U}(1)$ as a subgroup of $\operatorname{SL}_{2}(\mathbb{C})$ by the following map:

f:e^{it}\longmapsto\begin{pmatrix}e^{it}&\\ &e^{-it}\end{pmatrix}.

By Maschke’s Theorem for $\operatorname{U}(1)$ , Theorem 2.11.2(ii), $\rho\left(\begin{pmatrix}e^{it}&\\ &e^{-it}\end{pmatrix}\right)$ on $V$ can be diagonalised. Taking the derivative, we see that $\rho\left(\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}\right)$ can be diagonalised and hence so can $\rho(H)=-i\rho\left(\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}\right)$ .

In fact, the classification of irreducible representations of $\operatorname{U}(1)$ shows that $\rho(iH)$ has eigenvalues in $i\mathbb{Z}$ and so $\rho(H)$ has eigenvalues in $\mathbb{Z}$ . ∎

Remark 3.12.3.

The proof of the proposition is a instance of Weyl’s unitary trick. We turned the action of $H$ , which infinitesimally generates a non-compact one-parameter subgroup of $\operatorname{SL}_{2}(\mathbb{C})$ , into the action of the compact group $\operatorname{U}(1)$ infinitesimally generated by $iH$ . The action of this compact subgroup can be diagonalised.

The proposition does not hold for an arbitrary representation of the one-dimensional Lie algebra $\mathfrak{h}=\mathbb{C}H$ generated by $H$ . Namely, the map $zH\mapsto\begin{pmatrix}1&z\\ 0&1\end{pmatrix}$ cannot be diagonalised. It is implicitly the interaction of $H$ with the other generators $X$ and $Y$ which makes the proposition work.

Let $(\rho,V)$ be a finite-dimensional complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ . By Proposition 3.12.2 we get a decomposition

V=\bigoplus_{\alpha}V_{\alpha}

(3.12.4)

where each $V_{\alpha}$ is the eigenspace for $\rho(H)$ with eigenvalue $\alpha\in\mathbb{C}$ :

V_{\alpha}=\{v\in V\,|\,\rho(H){\bf v}=\alpha{\bf v}\}.

Definition 3.12.5.

(i)

Each $\alpha\in\mathbb{C}$ occurring in equation (3.12.4) is called a weight (more precisely, an $H$ -weight) for the representation $\rho$ .
(ii)

Each $V_{\alpha}$ is called a weight space for $\rho$ .
(iii)

The non-zero vectors in $V_{\alpha}$ are called weight vectors for $\rho$ .

Note that the weights corresponding to $V$ form a multiset, in other words a set with repeated elements. We say that a weight has multiplicity $n$ if it appears in the multiset $n$ times (in general, the multiplicity of a weight is the dimension of the weight space).

Example 3.12.6.

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

Example 3.12.7.

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

Example 3.12.8.

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 3.12.9.

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

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

and similarly

H({\bf e}_{1}\otimes{\bf e}_{-1})=H({\bf e}_{-1}\otimes{\bf e}_{1})=0,H({\bf e% }_{-1}\otimes{\bf e}_{-1})=-2{\bf e}_{-1}\otimes{\bf e}_{-1}

so that the weights are $\{-2,0,0,2\}$ .

Example 3.12.10.

Take $V=\operatorname{Sym}^{k}(\mathbb{C}^{2})$ . As $\{{\bf e}_{1},{\bf e}_{-1}\}$ is a basis of the standard representation, a set of basis vectors for $\operatorname{Sym}^{k}(\mathbb{C}^{2})$ is

\{{\bf e}_{1}^{a}{\bf e}_{-1}^{k-a}\,|\,0\leq a\leq k\}.

Given some arbitrary ${\bf e}_{1}^{a}{\bf e}_{-1}^{k-a}$ in the basis we calculate

	$\displaystyle H({\bf e}_{1}^{a}{\bf e}_{-1}^{k-a})$	$\displaystyle=a(H{\bf e}_{1}){\bf e}_{1}^{a-1}{\bf e}_{-1}^{k-a}+b(X{\bf e}_{-% 1}){\bf e}_{1}^{a}{\bf e}_{-1}^{{k-a}-1}$
		$\displaystyle=(2a-k){\bf e}_{1}^{a}{\bf e}_{-1}^{b}.$

Thus the weights are:

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

We will soon see an explanation for this pattern.

Proposition 3.12.11.

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

(i)

$\{\text{weights of }V\otimes W\}=\{\text{all sums of pairs of weights from }V% \text{ and }W\}$ .
(ii)

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

$\{\text{weights of }\bigwedge^{k}(V)\}=\{\text{sums of unordered `distinct' $k% $-tuples of weights of }V\}$ .

Proof.

Let ${\bf v}_{1},\ldots,{\bf v}_{n}$ be a basis of weight vectors of $V$ such that ${\bf v}_{i}$ has weight $\alpha_{i}$ , and let ${\bf w}_{1},\ldots,{\bf w}_{m}$ be similar for $W$ with weights $\beta_{i}$ . Then

	$\displaystyle H({\bf v}_{i}\otimes{\bf w}_{j})$	$\displaystyle=(H{\bf v}_{i}\otimes{\bf w}_{j})+{\bf v}_{i}\otimes(H{\bf w}_{j})$
		$\displaystyle=\alpha_{i}{\bf v}_{i}\otimes{\bf w}_{j}+{\bf v}_{i}\otimes\beta_% {j}{\bf w}_{j}$
		$\displaystyle=(\alpha_{i}+\beta_{j})({\bf v}_{i}\otimes{\bf w}_{j}).$

So $\{{\bf v}_{i}\otimes{\bf w}_{j}\}$ is a basis of $V\otimes W$ with the given weights. Parts (ii) and (iii) are left as an exercise (Problem 3.12.2). ∎

Example 3.12.12.

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 $\bigwedge^{2}(V)$ 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\}.

3.12.2. Exercises

Problem 30. Prove the rest of Proposition 3.12.11.