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Representation of sl(3,C) Dongseok KIM. Outline Adjoint representation of sl(3,C). Eigenspace decomposition with respect to H, the maximal abelian subalgebra.

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Presentation on theme: "Representation of sl(3,C) Dongseok KIM. Outline Adjoint representation of sl(3,C). Eigenspace decomposition with respect to H, the maximal abelian subalgebra."— Presentation transcript:

1 Representation of sl(3,C) Dongseok KIM

2 Outline Adjoint representation of sl(3,C). Eigenspace decomposition with respect to H, the maximal abelian subalgebra. Classify irreducible representations of sl(3,C). Tensor rules (weight, Brauer, honeycomb, hives) Invariant spaces, web spaces

3 The Lie algebra sl(3,C) is generated by

4 Adjoint representation of sl(3,C). We consider the general setup for the adjoint representation. First, we find L 0 =C L (h)=h. The set of all nonzero  2 h * for which L   0 is denoted by  and the elements of  are called the roots of sl(3,C) with respect to h. The decomposition sl(3,C)=h © ( ©  2  L  ) is called a Cartan decomposition or root space decomposition of sl}(3,C).

5 Since we can write h * =C[L 1, L 2, L 3 ]/(L 1 +L 2 +L 3 =0) where We find the Cartan decomposition of adjoint representation.

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7 Irreducible representation of sl(3,C) Let V be a finite dimensional irreducible representation of sl(3,C). First, we decompose V into eigenspaces of h. For the adjoint representation  was called a root, for other representation  is called a weight of V. A diagram of all weight with multiplicity is called a weight diagram. Then one can find a vector v 2 V  such that X ij (v)=0 which is called a highest weight vector of V and the corresponding weight is called highest weight of V. Then V is generated by the images of v under successive applications of three operatorsY ij.

8 Since sl(2,C) is immersed if we restrict actions of h to { X ij, Y ij, H ij }, we can see it reflects along three line =0. Thus given the highest weight, we can find the hexagonal (possibly triangular) region that all weights of V appear. The following figure shows how to find the region from the given highest weight .

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10 The next step will be how we deal with the interior of the region. In general, we find it by theorems by Freudenthal and Kostant. For sl(3,C), first the multiplicity of the boundary is 1. Next, we can shrink the boundary by one time actions of X ij, Y ij, if the boundary was hexagonal, the multiplicity of weight for the resulting polygon goes up by one, otherwise it stays the same. We continues the process until we find all multiplicity of weights.

11 Let 1 = L 1, 2 = - L 3. For finite dimensional representation, the highest weight must be in the region which can be expresses in non-negative integral linear combinations of 1 and 2. This region is called the Weyl chamber of sl(3,C). The following is a weight diagram of an irreducible representation of sl(3,C) of highest weight 4 1 +2 2, denoted by V 4   2 or V(4,2). The number in the weight diagram indicates the multiplicity of the weight.

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13 1)A given weight  in the Weyl chamber, there exists an irreducible representation of sl(3,C) with highest weight . Moreover  can be written as a linear combination of 1 and b 2, a 1 +b 2 where a, b are nonnegative integers. The irreducible representation is denoted by V(a,b) =V a 1 +b 2. 2) Any finite dimensional representation of sl(3,C) is of the form V a 1 +b 2.

14 Tensor Rule First we can use all weight diagram, for example we look at the defining representation of sl(3,C), V(1,0) which has three eigenvalues L 1, L 2 and L 3. Also one can find easily that its dual representation V(0,1) has three eigenvalues -L 1, -L 2 and -L 3. One can find that V(1,0) ­ V(0,1) had the following weight diagram which decompose into the adjoint representation and the trivial representation.

15 V(1,0) ­ V(2,1)

16 V(1,0) ­ V(2,1)  V(3,1) ©

17 V(1,0) ­ V(2,1)  V(3,1) © V(1,2) ©

18 Therefore we find that V(1,0) ­ V(2,1)  V(3,1) © V(1,2) © V(2,0) There is another way using the Brauer theorem

19 Honeycombs, V(1,0) is written by (0,1,1), V(2,1) by (0,2,3). Then we find all possible honeycombs. V(3,1) V(2,0) V(1,2)

20 Invariant space First we find few dimensions of invariants spaces –D–Dim (Inv (V(1,0) ­3 )) = 1, –D–Dim (Inv (V(0,1) ­3 )) = 1, –D–Dim (Inv (V ­ V * )) = 1, –D–Dim (Inv (V(1,0) ­ 2 ­ V(0,1) ­ 2 )) = 2. Let x, y, z be eigenvectors of V(1, 0) and let w = x ­ y ­ z – y ­ x ­ z + y ­ z ­ x – x ­ z ­ y + z ­ x ­ y – z ­ y ­ x. Then w 2 Inv (V(1, 0) ­3 ))

21 Webs for U_q(sl(3,C)) are generated by the following trivalent vertices with relations Webs for U_q(sl(3,C)) are generated by the following trivalent vertices with relations

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