6 SL₂ 6 SL₂ 6.2 Classification of representations of sl_2,C.

6.1 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
	$\displaystyle[X,Y]$	$\displaystyle=H.$

We 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 analyze the representation theory of $\mathfrak{sl}_{2,\mathbb{C}}$ .

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

Proposition 6.1.

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 6.2.

Let $(\rho,V)$ be a finite-dimensional complex-linear representation of $\mathfrak{sl}_{2,\mathbb{C}}$ . Then $\rho(H)$ is diagonalizable 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}\mapsto\begin{pmatrix}e^{it}&\\ &e^{-it}\end{pmatrix}.

By Maschke’s Theorem for $\operatorname{U}(1)$ , $\rho\left(\begin{pmatrix}e^{it}&\\ &e^{-it}\end{pmatrix}\right)$ on $V$ can be diagonalized. Taking the derivative, we see that $\rho\left(\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}\right)$ can be diagonalized 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 6.3.

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 $i H$ . The action of this compact subgroup can be diagonalized.

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 diagonalized. 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 LABEL:H-diag we get a decomposition

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

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

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

Definition 6.4.

1.

Each $\alpha\in\mathbb{C}$ occurring in equation (LABEL:weight-dec) 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 $\rho$ .

Example 6.5.

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

Example 6.6.

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 6.7.

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 6.8.

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 6.9.

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

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

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.

6.1.1 Highest weights

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

Lemma 6.10.

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 6.11.

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 6.12.

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

Proof.

Indeed, let $\alpha$ be the numerically greatest weight of $V$ (there must be one, as $V$ is finite-dimensional) and let $v$ be a weight vector of weight $\alpha$ . Then $X v$ has weight $\alpha+2$ by the fundamental weight calculation, so must be zero as $\alpha$ was maximal. ∎

Example 6.13.

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.