# Simple roots Lie algebra

If $\alpha$ and $\beta$ are simple roots, then $\alpha-\beta$ is not. This means that
$$E_{-\vec{\alpha}}|E_{\vec{\beta}}\rangle = 0$$
Now, according to the text I read, this means that $q$ in the formula
$$\frac{2\vec{\alpha}\cdot \vec{\mu}}{\vec{\alpha}^2}=-(p-q)$$
is zero, where $\vec{\mu}$ is a weight, and $p$ and $q$ are integers. I couldn't understand why $q=0$, if someone could explain to me.

## Answers and Replies

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

If $\alpha$ and $\beta$ are simple roots, then $\alpha-\beta$ is not. This means that
$$E_{-\vec{\alpha}}|E_{\vec{\beta}}\rangle = 0$$
Now, according to the text I read, this means that $q$ in the formula
$$\frac{2\vec{\alpha}\cdot \vec{\mu}}{\vec{\alpha}^2}=-(p-q)$$
is zero, where $\vec{\mu}$ is a weight, and $p$ and $q$ are integers. I couldn't understand why $q=0$, if someone could explain to me.
Eα is treated as a ladder operator, when it acts on a state of some weight say |μ> it adds to it and give something proportional to|μ+α>. E will do the opposite.

In general, ##(E_α)^{P+1}|μ> =C E_α|μ+pα>=0## and ##(E_{-α})^{q+1}|μ> =C E_α|μ-qα>=0## for positive integers p and q.

comparing to your equation,
##E_{-\vec{\alpha}}|μ\rangle = 0##, we have ##q=0##.

So q=0 is the lowest state. Thanks...

In this case only, not in general.