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Consider $A_1$ with basis $\{e,f,h\}$ and let $\phi$ be a three-dimension representation of $A_1$. Let $\phi(h)=\begin{pmatrix} 0 & 2 & 0 \\ 0 & -2 & 0 \\ 2 & -2 & 2 \end{pmatrix}$ and $\phi(f)=\begin{pmatrix} 0 & -1 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & -1 \end{pmatrix}$. I want to determine $\phi(e)$.

My first guess is to use that $\phi$ is a representation of a Lie algebra so

$\phi(h)=\phi([e,f])=[\phi(e),\phi(f)]=\phi(e)\phi(f)-\phi(f)\phi(e)$, so one get 9 equations determining $\phi(e)$. However there is not a unique solutions to these equation, so I get in doubt, if this is the correct way to do it. What is the approach here?

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1 Answer 1

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You can first find the weight spaces, i.e. the eigenspaces of $\phi(h)$. We have $$ v_2 = (0, 0, 1), \quad v_0 = (1, 0, -1), \quad v_{-2} = (-1, 1, 1) $$ forming an eigenbasis, where $\phi(h) v_2 = 2 v_2$, and $\phi(h) v_0 = 0 v_0$, and $\phi(h) v_{-2} = -2 v_{-2}$. We can summarise this by saying that $\phi(h) v_\lambda = \lambda v_\lambda$.

The operator $\phi(f)$ must decrease weights $2$ each time, meaning that $\phi(f) v_{\lambda}$ must be a multiple of $v_{\lambda + 2}$. Indeed, we have $\phi(f) v_2 = v_0$ and $\phi(f) v_0 = v_{-2}$ and $\phi(f) v_{-2} = 0$.

The operator $\phi(e)$ must increase weights by $2$ each time, so we already know that $\phi(e) v_2 = 0$, since $2$ is the largest weight in this representation. We will also have $$ \phi(e) v_0 = a v_2, \quad \phi(e) v_{-2} = b v_0,$$ for some complex numbers $a$ and $b$ which we need to determine. You can use the bracket to determine these, since we will have reduced that large matrix equation down to just a simple equation of numbers. For example, we have $$ -2 v_{-2} = \phi(h) v_{-2} = \phi(e) \phi(f) v_{-2} - \phi(f) \phi(e) v_{-2} = 0 - b v_{-2}$$ and hence $b = 2$.

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