I came across this question in Hall's book on "Lie Groups, Lie Algebra and Representations".

Let $\sigma: \mathfrak{sl}\left( 3, \mathbb{C} \right) \rightarrow \mathfrak{gl}\left( V \right)$ be a finite-dimensional irreducible representation with highest weight $\mu$. Let $v_0 \in V$ be a weight vector corresponding to the highest weight $\mu$. Then, the weight space corresponding to the weight $\mu - \alpha_1 - \alpha_2$ is atmost two-dimensional and is spanned by $\sigma \left( Y_1 \right) \sigma \left( Y_2 \right) v_0$ and $\sigma \left( Y_2 \right) \sigma \left( Y_1 \right) v_0$.

My strategy to approach this problem is to show that the weight space is exactly the span of the two given vectors. Since any more applications of $\sigma \left( X_i \right)$ or $\sigma \left( Y_i \right)$ would chance the weight, once I prove my claim, the result will be proved.

To see this result, one part is clear: $\text{span} \left\lbrace \sigma \left( Y_1 \right) \sigma \left( Y_2 \right) v_0, \sigma \left( Y_2 \right) \sigma \left( Y_1 \right) v_0 \right\rbrace \subseteq W$, where $W$ is the weight space corresponding to the weight $\mu-\alpha_1-\alpha_2$.

However, I am unable to prove the other way round. Starting with a vector $w \in W$, I was thinking to apply $\sigma \left( X_1 \right) \sigma \left( X_2 \right)$ and $\sigma \left( X_2 \right) \sigma \left( X_1 \right)$ to $w$ so that it comes to the weight space corresponding to the weight $\mu$. But since $\mu$ is the highest weight, we must have

$$\sigma \left( X_1 \right) \sigma \left( X_2 \right) w = \lambda_1 v_0,$$

$$\sigma \left( X_2 \right) \sigma \left( X_1 \right) w = \lambda_2 v_0.$$

However, after this, I am unsure of what to do. I tried applying $\sigma \left( Y_1 \right)$ and $\sigma \left( Y_2 \right)$ on both sides. But, I could not get anything from it. When using the commutation relations, the $X_1$ and $X_2$ do not disappear, which I want them to. Any help in this will be appreciated.

For reference, the symbols $\alpha_1 = \left( 2, -1 \right)$ and $\alpha_2 = \left( -1, 2 \right)$ are the positive simple roots of $\mathfrak{sl}\left( 3, \mathbb{C} \right)$.

For $\mathfrak{sl}\left( 3, \mathbb{C} \right)$, we have use the following basis

$$H_1 = \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{matrix} \right], \ H_2 = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix} \right],$$ $$X_1 = \left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right], \ X_2 = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right], \ X_3 = \left[ \begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right],$$ $$Y_1 = \left[ \begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right], \ Y_2 = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right], \ Y_3 = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix} \right].$$

Also, we have the following commutation relations:

$$[H_1, X_1] = 2X_1, [H_1, Y_1]=-2Y_1, [X_1, Y_1] = H_1, [H_2, X_2] = 2X_2, [H_2, Y_2] = -2Y_2, [X_2, Y_2] = H_2, [H_1, H_2] = 0,$$ $$[H_1, X_2] = -X_2, [H_1, X_3] = X_3, [H_1, Y_2] = Y_2, [H_1, Y_3] = -Y_3, [H_2, X_1] = -X_1, [H_2, X_3] = X_3, [H_2, Y_1] = Y_1,$$ $$[H_2, Y_3] = -Y_3, [X_1, X_2] = X_3, [X_1, Y_2] = 0, [X_1, Y_3] = -Y_2, [X_2, X_3] =0, [X_2, Y_1] = 0, [X_2, Y_3] = Y_1,$$ $$[X_3, Y_1] = -X_2, [X_3, Y_2] = X_1, [X_3, Y_3] = H_1 + H_2, [Y_1, Y_2] = -Y_3, [Y_1, Y_3] = 0, [Y_2, Y_3] = 0.$$


2 Answers 2


I don't know Hall's book, but here is how I would think about this problem: let $U=U(\mathfrak{sl}_3(\mathbf{C}))$ be the universal enveloping algebra. By the PBW theorem, this has basis given by all monomials

$$f_{12}^{m_{12}} f_1^{m_1} f_2^{m_2} h_1^{n_1} h_2^{n_2} e_{12}^{k_{12}} e_1^{k_1} e_2^{k_2},$$ where the exponents are non-negative integers and the relation with your notation is $f_1=Y_1$, $f_2=Y_2$, $f_{12}=Y_3$ and similarly for the $e$'s and $X$'s, $h_1=H_1$, and $h_2=H_2$. Now given a weight $\mu$ the highest weight module (or Verma module) corresponding to $\mu$ is the induced module $$M(\mu)=\mathrm{Ind}^U_{U_{\geq 0}}(\mathbf{C}_\mu),$$ where $U_{\geq 0}$ is the sub algebra generated by the $e$'s and $h$'s, with action on $\mathbf{C}$ given by $\mu(h_i)$ for $i=1,2$, and with $e$'s acting by $0$. By the PBW theorem $M(\mu)$ has basis given by all monomials $$f_{12}^{m_{12}} f_1^{m_1} f_2^{m_2},$$ with the weight of the given monomial given by $$\mu-(m_{12}+m_1)\alpha_1-(m_{12}+m_2) \alpha_2.$$ Now simply observe that the only monomials giving rise to your particular weight are for $m_{12}=1$, $m_1=0=m_2$ and $m_{12}=0$, $m_1=1=m_2$. This implies that the corresponding weight space in $M(\mu)$ is two-dimensional and spanned by $f_{12}$ and $f_1 f_2$ (use $[f_1,f_2]=f_{12}$ to get the spanning statement you need). Since every highest weight module of highest weight $\mu$ is a quotient of $M(\mu)$, this solves the problem.

A couple of remarks: first, this doesn't really have to do with finite-dimensionality, only with being generated by a highest-weight vector. Second, you don't really need the (more difficult) linear-independence part of the PBW theorem, only the statement that the given monomials are a spanning set.


In the case of $\mathrm{sl}(3,\mathbb{C})$, Hall constructs the representation with highest weight $\mu = (m,n)$ as the tensor product of $m$ copies of the standard representation of $\mathrm{sl}(3,\mathbb{C})$ on $\mathbb{C}^3$, and $n$ copies of the adjoint representation of $\mathrm{sl}(3,\mathbb{C})$. This can be done without invoking Verma modules.

Using the notation of Section 6.5 (at least of my version), which is based on the example with the highest weight (1,1), Hall notes that the standard representation has a basis $\{e_1,e_2,e_3\}$ such that $$Y_1e_1 = e_2, Y_2e_2 = e_3,$$ and every other combo gives a zero.

Hall also notes that the adjoint representation has a basis $\{f_1,f_2,f_3\}$ such that $$Y_2f_1=f_2, Y_1f_2=f_3,$$ and every other combo gives a zero.

The highest weights of these representations are given by $e_1$ and $f_1$ respectively. Following the proof of Proposition 6.17, it is easily verified that $w := e_1 \otimes \ldots \otimes e_1 \otimes f_1 \otimes \ldots \otimes f_1$, with $m$ copies of $e_1$ and $n$ copies of $f_1$, is a weight vector with weight $(m,n)$ that is annihilated by the $X_i$.

Key point: Let $Y_{i_1}\ldots Y_{i_k}$ be some string of $r$ $Y_1$'s and $s$ $Y_2$'s, though in some order. Then $Y_{i_1}\ldots Y_{i_k} w$ is a weight vector with weight $(m,n) -r\alpha_1 -s\alpha_2$. In particular, the dimension of the weight space with this weight is at most $\binom{r+s}{r}$, which is the number of distinct ways of ordering $r$ copies of $Y_1$ with $s$ copies of $Y_2$. In your case, $r=s=1$, we see that there are at most $2$ distinct ways.


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