Decomposition of irrep. of G_2 into irreps. of A_2

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Hello everybody,

this is the first thread of mine. I try to recover the decomposition of
the fundamental representation (the 7) of G2 into irreducible representations of
A_2. It is given by

7 = 3 + \bar {3}[\tex] + 1<br /> <br /> It maid be that I didn&#039;t understand the procedure, founded by Dynkin, to use the <br /> extended Dynkin diagram. <br /> <br /> o<u>==</u>o--o<br /> a1 a2 ax<br /> <br /> Where the last root is the extension. The weights of the 7 are :<br /> <br /> {1, 2}, <br /> {1, 1},<br /> {0, 1}, <br /> {0, 0}, <br /> {0, -1}, <br /> {-1, -1}, <br /> {-1, -2}<br /> <br /> I&#039;am using the Cartan subalgebra {a2,ax} to embed the A_2 in G_2. I think that this is<br /> the point where I do a mistake? Maybe someone could help me with the decompostion.<br /> <br /> Thanks for reading<br /> <br /> Timb00
 
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You can take a look at Fulton and Harris book "Representation Theory", lecture 22. It is about the Lie algebra \mathfrak g_2 and includes the branching law that you ask about.
 
Hey Guy's,

thanks for reading my thread. I found my mistake. It was just the wrong normalization.
If someone is interested I will make a sketch how I compute the weight projector.

For the A_2 I take the Cartan sub algebra to be generated out of {a'1,a'2}={4/3*a2,4/3*ax}. The factor of 4/3 is because of the normalization. Now it is possible to get two independent equations :

(m_1 a1 + m_2 a2) (a'1) =4/3(m_1 a1 + m_2 a2)(a2)=(n_1 a'_1 + n_2 a'_2)(a'1)
<=> 4/3(m_2 1/4-m_1/8) = n_1 1/3 - n_2 1/6

and

(m_1 a1 + m_2 a2) (a'2) =4/3(m_1 a1 + m_2 a2)(ax)=(n_1 a'_1 + n_2 a'_2)(a'2)
<=> 4/3(m_2 1/4) = n_2 1/3 - n_1 1/6

In matrix notation one finds :

A(m_1,m_2)^T = B (n_1,m_2)^T <=> P (m_1,m_2)^T = (n_1,m_2)^T

=> P = B^-1A

If one acts with P on the weights one findes that the set of G_2 weights decompose into
a 3 + {\bar 3} +1 .

Thanks, timb00.

P.s : hey martinbn, thanks for your answer. I saw it after I wrote the text. But I will have
a look into the Fulton.
 
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