4.20. Lecture 20

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1. (i)
   ​

   Let

   $$W_{n}=\{\rho(Y_{m})\rho(Y_{m-1})\ldots\rho(Y_{1}){\bf v}\,|\,n\geq m\geq 0,Y_{i}\in\{E_{21},E_{32}\}\}.$$

   Then

   $$W=\bigcup_{n=0}^{\infty}W_{n}.$$

   Firstly, it is clear that $\rho(E_{21})$ and $\rho(E_{32})$ take $W_{n}$ to $W_{n+1}$, and so preserve $W$. Since

   $$\rho(E_{31})=\rho([E_{32},E_{21}])=[\rho(E_{32}),\rho(E_{21})]$$

   we see that $\rho(E_{31})$ also preserves $W$.

   Secondly, every $\rho(Y_{m})\rho(Y_{m-1})\ldots\rho(Y_{1}){\bf v}$ is a weight vector (by the fundamental weight calculation, Lemma 4.19.1) and so an eigenvector for all $\rho(H)$, $H\in\mathfrak{h}$. Thus $\rho(H)$ preserves each $W_{n}$ (and hence also $W$).

   Finally, we show that $\rho(E_{12})$ preserves $W_{n}$. A similar proof then applies for $\rho(E_{23})$, and then $\rho(E_{13})$ preserves $W$ by the same argument as for $\rho(E_{31})$. We prove the statement for $\rho(E_{12})$ by induction on $n$.

   For $n=0$, $W_{0}=\left\langle{\bf v}\right\rangle$. Since ${\bf v}$ is a highest weight vector, $\rho(E_{12}){\bf v}=0$ and so $\rho(E_{12})(W_{0})\subseteq W_{0}$.

   Suppose that the claim is true for $n$. Consider ${\bf w}=\rho(Y_{n+1})\ldots\rho(Y_{1}){\bf v}\in W_{n+1}$ with $Y_{i}\in\{E_{21},E_{32}\}$. We must show that $\rho(E_{12}){\bf w}\in W_{n+1}$. Suppose first that $Y_{n+1}=E_{21}$. Then, as $[E_{12},E_{21}]=H_{12}\in\mathfrak{h}$, we have

   	$\displaystyle\rho(E_{12}){\bf w}$	$\displaystyle=\rho(E_{12})\rho(E_{21})\rho(Y_{n})\ldots\rho(Y_{1}){\bf v}$
   		$\displaystyle=\rho(E_{21})\rho(E_{12})\rho(Y_{n})\ldots\rho(Y_{1}){\bf v}+\rho(H_{12})\rho(Y_{n})\ldots\rho(Y_{1}){\bf v}$
   		$\displaystyle\in\rho(E_{21})\rho(E_{12})W_{n}+\rho(H_{12})W_{n}$
   		$\displaystyle\subseteq\rho(E_{21})W_{n}+W_{n}$
   by the induction hypothesis and the fact that $W_{n}$ is preserved by $\mathfrak{h}$
   		$\displaystyle\subseteq W_{n+1}+W_{n}$
   		$\displaystyle=W_{n+1},$

   as required. The proof in the case $Y_{n+1}=E_{32}$ is similar, using that $[E_{12},E_{32}]=0$.

2. (ii)
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   Note that if $\beta$ the weight of $\rho(Y_{n})\rho(Y_{m-1})\ldots\rho(Y_{1}){\bf v}$, with $n$ and $Y_{i}$ as in the lemma, then a calculation using the fundamental weight calculation, as in the proof of lemma 4.19.7, shows that

   $$l(\beta)=l(\alpha)-n$$

   and so $\beta\neq\alpha$ if $n>0$. Since these vectors span $W$, ${\bf v}$ is the unique (up to scalar) weight vector in $W$ of weight $\alpha$.

3. (iii)
   ​

   Suppose that $W$ is reducible. By complete reducibility (Theorem 3.14.7) we have

   $$W=U\oplus U^{\prime}$$

   for $U,U^{\prime}$ non-zero proper subrepresentations of $W$. We must have ${\bf v}={\bf u}+{\bf u}^{\prime}$ for unique ${\bf u}\in U$, ${\bf u}^{\prime}\in U^{\prime}$. The unicity implies that ${\bf u}$ and ${\bf u}^{\prime}$ are both weight vectors of weight $\alpha$ so, by part (ii), either ${\bf u}=0$ or ${\bf u}^{\prime}=0$. Without loss of generality let ${\bf v}={\bf u}\in U$. But then all $\rho(Y_{n})\ldots\rho(Y_{1}){\bf v}\in U$ as $U$ is a subrepresentation, so $W=U$ contradicting that $U$ is a proper subrepresentation.

∎

First we show the existence. Let $a,b\in\mathbb{Z}_{\geq 0}$. Consider

This has a highest weight vector ${\bf v}={\bf e}_{1}^{a}\otimes({\bf e}_{3}^{*})^{b}$ of weight $aL_{1}-bL_{3}$. Let $W$ be the representation generated by ${\bf v}$. Then $W$ is irreducible by 4.20.3(iii), and has a highest weight vector ${\bf v}$ of weight $aL_{1}-bL_{3}$. Thus we can take $V^{(a,b)}=W$.

Next we show the uniqueness. Suppose that $V,W$ are two irreducible representations with highest weight vectors ${\bf v}$ and ${\bf w}$, respectively, of weight $aL_{1}-bL_{3}$. Let $U\subseteq V\oplus W$ be the representation generated by ${\bf u}=({\bf v},{\bf w})$. Then $U$ is irreducible by 4.20.3(iii). The projection $V\oplus W\to V$ sending $({\bf v}^{\prime},{\bf w}^{\prime})$ to ${\bf v}^{\prime}$ restricts to a homomorphism $U\to V$ which sends ${\bf u}$ to ${\bf v}$. This is therefore a non-zero homomorphism between irreducible representations, and so must be an isomorphism. Thus $U\cong V$. Similarly $U\cong W$, and so $V\cong W$ as required. ∎

This is Problem 4.7. ∎