5.1 Finite dimensional representations of sl(2, C)

TLDR

Any irreducible finite dimensional representation of $s{l}_2(\mathbb{C})$, or any finite dimensional $s{l}_2(\mathbb{C})$-module, is isomorphic to a $({\pi }_m,V_m)$, which is uniquely determined by its dimension. Furthermore, any finite dimensional $s{l}_2(\mathbb{C})$-module can be written as a direct sum of irreducible $s{l}_2(\mathbb{C})$-submodules.

Irreducible representations of $s{l}_2(\mathbb{C})$

Notice that

$$s{l}_2(\mathbb{C})=\mathbb{C}\left[\begin{array}{cc}1& \\ & -1\end{array}\right]\oplus \mathbb{C}\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\oplus \mathbb{C}\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right].$$

Define $h=\left[\begin{array}{cc}1& \\ & -1\end{array}\right]$, $e=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ and $f=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$.

Lemma

Let $V$ be an $s{l}_2(\mathbb{C})$-module. Suppose $v\in V$ satisfies $hv=kv$. Then:

• $h{e}^{i}v=(2i+k)e^{i}v$, $h{f}^{i}v=(-2i+k)f^{i}v$.
• When $ev=0$, then $e{f}^{i}v=i(k-i+1)f^{i-1}v$.
• When $ev=0$ and $f^{m}v\ne 0$, $f^{m+1}v=0$, the space $V:=\mathrm{span}\{v,fv,\cdots ,f^{m}v\}$ is a $(m+1)$-dim irreducible $s{l}_2(\mathbb{C})$-module and $k=m$.

\begin{proof}i) & ii) By induction.

iii) By i) and ii), it is easy to verify $V$ a $s{l}_2(\mathbb{C})$-module. For any element $v\in V$, $v={\sum }_{i=0}^m{a}_i{f}^i$, then $f^{k}v=c{f}^{m}v$ for some $k\in \mathbb{Z}$. And so ${\mathrm{sl}}_2(\mathbb{C})$ acting on $v$ generates the whole space $V$, i.e. $V$ is irreducible. \end{proof}

Proposition

Let ${\pi }_m(m\ge 0)$ be the $s{l}_2(\mathbb{C})$-module given by $V_m={\mathbb{C}}^{m+1}=\mathrm{span}\{v_0,\cdots ,v_m\}$, where

$$\begin{array}{rl}& {\pi }_m(h)v_i:=(m-2i)v_i,\\ & {\pi }_m(f):v_0\to v_1,v_1\to v_2,\cdots ,v_{m-1}\to v_m,v_m\to 0,\\ & {\pi }_m(e):v_i\to i(m-i+1)v_{i-1}.\end{array}$$

These are the only finite dimensional irreducible $s{l}_2(\mathbb{C})$-modules.

\begin{proof}Let $(V,\pi )$ be a finite dimensional irreducible $s{l}_2(\mathbb{C})$-modules. We will show there is a $m\in {\mathbb{Z}}_{\ge 0}$ s.t. $\pi \cong {\pi }_m$. Let $v_0$ be an eigenvector of $\pi (h)$ with maximal eigenvalue $\lambda$. Since $\pi (h)\pi (e)v=(\lambda +2)\pi (e)v$, there must be $\pi (e)v_0=0$. On the other hand, $\pi (h)$ has only finitely many eigenvalues and $\pi (h){\pi }^i(f)v_0=(\lambda -2i){\pi }^i(f)v_0$, it follows that there exists $m\in {\mathbb{Z}}_{\ge 0}$ s.t. $\pi (f{)}^m{v}_0\ne 0$ and $\pi (f{)}^{m+1}{v}_0=0$. Then

$$\mathrm{span}\{v_0,\pi (f)v_0,\cdots ,\pi (f{)}^m{v}_0\}$$

is an irreducible $s{l}_2(\mathbb{C})$-module of dimension $m+1$. By irreducibility of $\pi$, there is $V=\mathrm{span}\{v_0,\pi (f)v_0,\cdots ,\pi (f{)}^m{v}_0\}$ and so $\pi \simeq {\pi }_m$. \end{proof}

Completely reducibility

We want to show that finite dimensional representations of $s{l}_2(\mathbb{C})$ are completely reducible. To show it, we need the following criterion.

The criterion for completely reducibility

the criterion for complete reducibility

Suppose $g$ is finite dimensional complex Lie algebra with $g=[g,g]$. Then the followings are equivalent:

• For any finite dimensional $g$-modules $N\subseteq M$, there exists a complementary $g$-submodule $P$ of $M$ such that $M\simeq N\oplus P$.
• For any finite dimensional $g$-modules $N\subseteq M$ with $\dim M/N=1$, there is $M\simeq N\oplus \mathbb{C}$.
• For any finite dimensional $g$-modules $N\subseteq M$ with $N$ irreducible and $\dim M/N=1$, $M\simeq N\oplus \mathbb{C}$.

\begin{proof}i) → ii) → iii) is trivial.

To show iii) → ii), we do induction on $N$. When $\dim N=1$, $N$ is irreducible and we apply iii). Assume ii) holds when $\dim N\le n$. Now suppose $\dim N=n+1$. If $N$ is irreducible, we merely apple iii). Otherwise, choose an irreducible $g$-submodule $N_0⊊N$. Consider $N/N_0\hookrightarrow M/N_0$. Applying induction hypothesis, one get $M/N_0\simeq N/N_0\oplus \mathbb{C}$ where the summand $\mathbb{C}$ corresponds $P/N_0$, where $P$ is a $g$-submodule containing $N_0$. Applying iii) to $P\supseteq N_0$, we get $P=N_0\oplus P^{\prime }$ with $P^{\prime }\simeq \mathbb{C}$. Now $M=N\oplus P^{\prime }$.

To show ii) → i), we need to find a $g$-module homomorphism $\varphi :M\to N$ such that $\varphi {|}_N={\mathrm{id}}_N$. Consider the vector space ${\mathrm{Hom}}_{\mathbb{C}}(M,N)$ containing ${\mathrm{Hom}}_g(M,N)$, which is a natural $g$-module via the action

$$X\circ \varphi (m)=X(\varphi (m))-\varphi (X(m)),\varphi \in {\mathrm{Hom}}_{\mathbb{C}}(M,N),X\in g,m\in M.$$

Clearly, an element $\varphi \in {\mathrm{Hom}}_{\mathbb{C}}(M,N)$ belongs to ${\mathrm{Hom}}_g(M,N)$ iff $g\circ \varphi =0$.

Now consider

$$\begin{array}{rl}\pi :{\mathrm{Hom}}_{\mathbb{C}}(M,N)& \to {\mathrm{Hom}}_{\mathbb{C}}(N,N),\\ \varphi & \mapsto \varphi {|}_N.\end{array}$$

There are

$$0\hookrightarrow \mathbb{C}\mathrm{id}{|}_N\hookrightarrow {\mathrm{Hom}}_{\mathbb{C}}(N,N)$$

and

$${\pi }^{-1}(0)\hookrightarrow {\pi }^{-1}(\mathbb{C}\mathrm{id}{|}_N)\hookrightarrow {\mathrm{Hom}}_{\mathbb{C}}(M,N).$$

So by ii) we have ${\pi }^{-1}(\mathbb{C}{\mathrm{id}}_N)={\pi }^{-1}(0)\oplus \mathbb{C}\varphi$, where $\mathbb{C}\varphi$ is a $1$-dimensional submodule complementary to ${\pi }^{-1}(0)$. We can make $\pi (\varphi )=\mathrm{id}{|}_N$. Also, $\mathbb{C}\varphi \subseteq {\mathrm{Hom}}_{\mathbb{C}}(M,N)$ as a $1$-dimensional $g$-module must be trivial by $g=[g,g]$. It follows that $\varphi \in {\mathrm{Hom}}_g(M,N)$.

Notes on the class is attached here. \end{proof}

Main theorem

Now we can prove the following theorem.

Theorem

Every finite dimensional representation $(\pi ,V)$ of $s{l}_2(\mathbb{C})$ is completely reducible.

\begin{proof}By the criterion for complete reducibility, it suffices to show for any finite dimensional $g$-module $N\subseteq M$ with $N$ irreducible and $\dim M/N=1$, there is $M\simeq N\oplus \mathbb{C}$.

Observation: denotes the action of $s{l}_2(\mathbb{C})$ on $M$ by $\pi$. If $\varphi$ is in the algebra generated by $\pi (s{l}_2(\mathbb{C}))$ and commutes with $\pi (s{l}_2(\mathbb{C}))$, then $\varphi$ is a $s{l}_2(\mathbb{C})$-module homomorphism. By Schur’s lemma, $\varphi$ acts on $N$ by scalar. (Schur’s lemma is proved by consider the kernel of $\varphi -\lambda I$, by algebraically closed there exists a $\lambda$ such that $\ker (\varphi -\lambda I)\ne \varnothing$ and so $\ker (\varphi -\lambda I)=N$.)

We may distinguish the target $1$-dimensional complementary subspace $P$ using eigenvalues of $\varphi$.

Consider $\Omega =ef+h(h-2)/4=fe+h(h+2)/4$. Then $\Omega$ commutes with $h,e,f$. Since $N$ is irreducible, it is $({\pi }_k,V_k)$ for some $k\in {\mathbb{Z}}_{\ge 0}$. $\pi (\Omega )$ acts on $V_k$ by some scalar and it’s scalar can be computed using the highest weight vector $v_t$. Since

$$\pi (\Omega )v_t=(\pi (f)\pi (e)+\pi (h)\pi (h+2)/4)v_t=k(k+2)/4v_t,$$

$\pi (\Omega )$ acts on $N$ by $k(k+2)/4$.

Case 1: If $k>0$, in this situation, $\pi (\Omega )$ acts on $N$ by a non-zero scalar $k(k+2)/4$. On the other hand, $M/N$ is a trivial $s{l}_2(\mathbb{C})$-module by $s{l}_2(\mathbb{C})=[s{l}_2(\mathbb{C}),s{l}_2(\mathbb{C})]$ and $\overline{\pi }(\Omega )$ acts by the zero vector. Whence $\pi (\Omega )$ has an eigenvalue $0$. Thus the $0$-eigenspace $U$ of $\pi (\Omega )$ must be $1$-dimensional. Because $\pi (\Omega )$ commutes with $\pi (s{l}_2(\mathbb{C}))$, $\pi (\Omega )\pi (s{l}_2(\mathbb{C}))v=\pi (s{l}_2(\mathbb{C}))\pi (\Omega )v=0$ for $0$-eigenvector $v$, $U$ is $s{l}_2(\mathbb{C})$-invariant. Thus $M\simeq N\oplus U$.

Case 2: If $k=0$, $N$ is the $1$-dimensional trivial $s{l}_2(\mathbb{C})$-module. Choosing a basis of $N$ and extending it to a basis of $M$, then the matrix representation of $\pi$ is

$$s{l}_2(\mathbb{C})\to \left\{\left[\begin{array}{cc}0& \ast \\ & 0\end{array}\right]\right\}.$$

Because $s{l}_2(\mathbb{C})=[s{l}_2(\mathbb{C}),s{l}_2(\mathbb{C})]$, $\pi (s{l}_2(\mathbb{C}))=0$. Hence any complementary subspace of $N$ is $s{l}_2(\mathbb{C})$-invariant and so we find $P$ s.t. $M\simeq N\oplus P$. \end{proof}