Determinie the multiplicity of a root in $\mathfrak{g}(A)$ where $2\times 2$-matrix $A$ given (or arbitrary)

Question

I'm trying to solve the exercises on Infinite-dimensional Lie algebras by Victor G.Kac.

Exercise 1.6.:

Let $A = \begin{pmatrix}2&-3\\-3&2\end{pmatrix}$. Show that $\operatorname{mult} (2\alpha_1+3\alpha_2) =2$ in $\mathfrak{g}(A)$. Show that for an arbitrary $2\times 2$-matrix $A$, $\operatorname{mult} (2\alpha_1+3\alpha_2) \leq 2$; find the conditions when it is $= 2$.

Here $\operatorname{mult} \alpha$ is the dimension of the corrsponding root space $\mathfrak{g}_{\alpha}$ in $\mathfrak{g}(A)$.

The root space $\mathfrak{g}_{2\alpha_1+3\alpha_2}$ is spanned by $x_1=[[[[e_1,e_2],e_2],e_2],e_1]$ and $x_2=[[[[e_1,e_2],e_1],e_2],e_2]$. So we have $\operatorname{mult} (2\alpha_1+3\alpha_2) \leq 2$ for an arbitrary $2\times 2$-matrix $A$.

I don't know how to determine the multiplicity with the Cartan matrix. I have tried to show that the ideal generated by $k_1x_1+k_2x_2(k_1,k_2\in\mathbb{C})$ in $\tilde{\mathfrak{g}}(A)$ intersects $\mathfrak{h}$ trivially, but the calculation seems to be too much heavy.

Answer 1

By Exercise 1.5, the root space $\mathfrak{g}_{2\alpha_1+3\alpha_2}$ is spanned by $x_1=[[[[e_1,e_2],e_2],e_2],e_1]$ and $x_2=[[[[e_1,e_2],e_1],e_2],e_2]$. So we have $\operatorname{mult} (2\alpha_1+3\alpha_2) \leq 2$ for an arbitrary $2\times 2$-matrix $A$.

Now consider the conditions when it is $= 2$. Note that \begin{align*} [[e_1,e_2],f_1]&=a_{12}e_2,\\ [[e_1,e_2],f_2]&=-a_{21}e_1,\\ [[[e_1,e_2],e_2],f_2]&=-(2a_{21}+a_{22})[e_1,e_2],\\ [[[e_1,e_2],e_1],f_1]&=-(a_{11}+2a_{12})[e_1,e_2],\\ [[[[e_1,e_2],e_2],e_2],f_2]&=-3(a_{21}+a_{22})[[e_1,e_2],e_2],\\ [[[[e_1,e_2],e_1],e_2],f_1]&=-(a_{11}+2a_{12})[[e_1,e_2],e_2],\\ [[[[e_1,e_2],e_1],e_2],f_2]&=-(2a_{21}+a_{22})[[e_1,e_2],e_1],\\ [k_1x_1+k_2x_2,f_1]&=-(k_1(a_{11}+a_{12})+k_2(a_{11}+2a_{12}))[[[e_1,e_2],e_2],e_2],\\ [k_1x_1+k_2x_2,f_2]&=-(3k_1(a_{21}+a_{22})+k_2(4a_{21}+3a_{22}))[[[e_1,e_2],e_1],e_2]. \end{align*}

Thus by Lemma 1.5, the conditions when it is $= 2$ are

• $a_{12}\neq 0$ or $a_{21}\neq 0$ (hence $[e_1,e_2]\neq 0$);
• $a_{11}+2a_{12},~2a_{21}+a_{22},~a_{21}+a_{22}$ are all not equal to zero (hence $[[[e_1,e_2],e_2],e_2],~[[[e_1,e_2],e_1],e_2]\neq 0$);
• $\det\begin{pmatrix} a_{11}+a_{12}&a_{11}+2a_{12}\\3(a_{21}+a_{22})&4a_{21}+3a_{22} \end{pmatrix}\neq 0$, i.e. $\det\begin{pmatrix}a_{11}& a_{12}\\2a_{21}+3a_{22}& a_{21}\end{pmatrix}\neq 0$ (hence $x_1$ and $x_2$ are linearly independent).

So we have $\operatorname{mult} (2\alpha_1+3\alpha_2) =2$ in $\mathfrak{g}(A)$ if $A = \begin{pmatrix}2&-3\\-3&2\end{pmatrix}$.