Highest weight of representations of Lie Algebras

[physicist_2be]

Hello there,

Given a Lie Algebra $\mathfrak{g}$, its Cartan Matrix $A$ and a finite representation $R$, is there a way of determining its highest weight $\Lambda$ in a simple way?

In my course, we consider $\mathfrak{g}=A_2= \mathfrak{L}_{\mathbb{C}}(SU(3))$. It is stated that the highest weight of the fundamental representation has Dynkin labels $\Lambda = (1,0)$ and the highest weight of the adjoint representation has Dynkin labels $\Lambda = (1,1)$. Why is it so? From there, I can work out the other roots by removing weights given by the Cartan Matrix but it is of no use if I can't compute the highest weight in the first place.

Taking an example, let $\mathfrak{g}=B_2= \mathfrak{L}_{\mathbb{C}}(SO(5))$. How do I work out the highest weight for the fundamental and adjoint representation?

Thanks in advance!

 

[fresh_42]

Hello there,

Given a Lie Algebra $\mathfrak{g}$, its Cartan Matrix $A$ and a finite representation $R$, is there a way of determining its highest weight $\Lambda$ in a simple way?

In my course, we consider $\mathfrak{g}=A_2= \mathfrak{L}_{\mathbb{C}}(SU(3))$. It is stated that the highest weight of the fundamental representation has Dynkin labels $\Lambda = (1,0)$ and the highest weight of the adjoint representation has Dynkin labels $\Lambda = (1,1)$. Why is it so? From there, I can work out the other roots by removing weights given by the Cartan Matrix but it is of no use if I can't compute the highest weight in the first place.

Taking an example, let $\mathfrak{g}=B_2= \mathfrak{L}_{\mathbb{C}}(SO(5))$. How do I work out the highest weight for the fundamental and adjoint representation?

Thanks in advance!

You need to say semisimple or simple Lie algebra!
Here's an example of how to construct the root spaces from the Dynkin diagram, or the Cartan matrix:
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-structures/

I think highest weight is in general a bit more complicated. $\mathfrak{sl}(2)$ is easy, but with more than one strain of roots it becomes more complex. I would look it up in books, e.g. Humphreys.