Simple roots Lie algebra

  • Thread starter spookyfish
  • Start date
  • #1
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If [itex]\alpha [/itex] and [itex]\beta [/itex] are simple roots, then [itex]\alpha-\beta [/itex] is not. This means that
[tex]
E_{-\vec{\alpha}}|E_{\vec{\beta}}\rangle = 0
[/tex]
Now, according to the text I read, this means that [itex]q [/itex] in the formula
[tex]
\frac{2\vec{\alpha}\cdot \vec{\mu}}{\vec{\alpha}^2}=-(p-q)
[/tex]
is zero, where [itex]\vec{\mu} [/itex] is a weight, and [itex]p[/itex] and [itex]q [/itex] are integers. I couldn't understand why [itex]q=0 [/itex], if someone could explain to me.
 

Answers and Replies

  • #2
18,488
8,326
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?
 
  • #3
1,024
32
If [itex]\alpha [/itex] and [itex]\beta [/itex] are simple roots, then [itex]\alpha-\beta [/itex] is not. This means that
[tex]
E_{-\vec{\alpha}}|E_{\vec{\beta}}\rangle = 0
[/tex]
Now, according to the text I read, this means that [itex]q [/itex] in the formula
[tex]
\frac{2\vec{\alpha}\cdot \vec{\mu}}{\vec{\alpha}^2}=-(p-q)
[/tex]
is zero, where [itex]\vec{\mu} [/itex] is a weight, and [itex]p[/itex] and [itex]q [/itex] are integers. I couldn't understand why [itex]q=0 [/itex], 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##.
 
  • #4
53
0
So q=0 is the lowest state. Thanks...
 
  • #5
1,024
32
In this case only, not in general.
 

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