6 SL₂ 6.1 Weights 6.3 Decomposing representations

6.2 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 6.14.

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).

(6.1)

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 6.15.

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

Corollary 6.16.

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 6.2.

Theorem 6.17.

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 6.14 its weight is a nonnegative integer $n$ , and by Corollary 6.16 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
	$\displaystyle\rho(X)$	$\displaystyle=\begin{pmatrix}\hidden@noalign{}\hfil\textstyle 0&n&&&&\\ \hidden@noalign{}\hfil\textstyle&0&2(n-1)&&&\\ \hidden@noalign{}\hfil\textstyle&&0&3(n-2)&&\\ \hidden@noalign{}\hfil\textstyle&&&\ddots&\ddots&\\ \hidden@noalign{}\hfil\textstyle&&&&\ddots&n\\ \hidden@noalign{}\hfil\textstyle&&&&&0\end{pmatrix}.$

Theorem 6.18.

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 $\mathfrak{sl}_{2,\mathbb{C}}$ . If $V$ is a representation of $\operatorname{SL}_{2}(\mathbb{C})$ , then its derivative will be isomorphic to $\operatorname{Sym}^{n}(\mathbb{C}^{2})$ . Since $\operatorname{SL}_{2}(\mathbb{C})$ is connected, this implies that $V\cong\operatorname{Sym}^{n}(\mathbb{C}^{2})$ . ∎

6.2 Classification of representations of sl2,C.

Lemma 6.14.

Proof.

Remark 6.15.

Corollary 6.16.

Proof.

Theorem 6.17.

Proof.

Theorem 6.18.

Proof.

6.2 Classification of representations of sl_2,C.