Why is q=0 for Simple Roots in Lie Algebras?

[spookyfish]

If \alpha and \beta are simple roots, then \alpha-\beta is not. This means that
<br /> E_{-\vec{\alpha}}|E_{\vec{\beta}}\rangle = 0<br />
Now, according to the text I read, this means that q in the formula
<br /> \frac{2\vec{\alpha}\cdot \vec{\mu}}{\vec{\alpha}^2}=-(p-q)<br />
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.

 

[Greg Bernhardt]

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?

 

[andrien]

If \alpha and \beta are simple roots, then \alpha-\beta is not. This means that
<br /> E_{-\vec{\alpha}}|E_{\vec{\beta}}\rangle = 0<br />
Now, according to the text I read, this means that q in the formula
<br /> \frac{2\vec{\alpha}\cdot \vec{\mu}}{\vec{\alpha}^2}=-(p-q)<br />
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.

E_α 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_-α 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 $q=0$.

 

[spookyfish]

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

 

[andrien]

In this case only, not in general.