Decomposition of irrep. of G_2 into irreps. of A_2

  • Thread starter timb00
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  • #1
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Main Question or Discussion Point

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 + [tex]\bar {3}[\tex] + 1

It maid be that I didn't understand the procedure, founded by Dynkin, to use the
extended Dynkin diagram.

o==o--o
a1 a2 ax

Where the last root is the extension. The weights of the 7 are :

{1, 2},
{1, 1},
{0, 1},
{0, 0},
{0, -1},
{-1, -1},
{-1, -2}

I'am using the Cartan subalgebra {a2,ax} to embed the A_2 in G_2. I think that this is
the point where I do a mistake? Maybe someone could help me with the decompostion.

Thanks for reading

Timb00
 

Answers and Replies

  • #2
martinbn
Science Advisor
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You can take a look at Fulton and Harris book "Representation Theory", lecture 22. It is about the Lie algebra [tex]\mathfrak g_2[/tex] and includes the branching law that you ask about.
 
  • #3
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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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