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How can the multiplication of spinor representations (of $SO(8)$) $8_+ \otimes 8_-$ be decomposed into $8_v \oplus 56_v$? Where can I read more about the decomposition rule of different representations?

Thank you.

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Appendix B of Polchinski Vol. 2 has a good discussion of this. –  Prahar Jun 29 '13 at 21:57

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up vote 8 down vote accepted

The given representations are not very large, which makes the computation from first principles not very cumbersome. Here I follow Slansky. It should be emphasized, that there exist methods (combinatorial and other) which reduce the computational complexity of some of the following steps, but they require more advanced representation theory than the Cartan-Weyl theory which is going to be used.

The highest weights of the representations under considerations are (Slansky: table 36):

$8_v: [1, 0, 0, 0]$

$8_+: [0, 0, 1, 0]$

$8_-: [0, 0, 0, 1]$

$56_v: [0, 0, 1, 1]$

The Cartan matrix of $SO(8)$ is given by: (Slansky: table 6)

$$\begin{pmatrix} 2& -1 & 0& 0 \\ -1& 2&-1 & -1\\ 0 & -1 &2 &0 \\ 0& -1 & 0 & 2 \end{pmatrix}$$

The weight diagrams of the spinor representations can be constructed using the method described by Slansky: page 31, which may be summarized as follows: Starting from the highest weight, if the $i$-th component of the weight is a positive integer $n_i$, then the primitive weight $\alpha_i$ is subtracted $n_i$ times. If the component is zero or negative, no subtraction is performed. This procedure is repeated until a negative weight is reached (the lowest weight).

This procedure does not give the multiplicity of the weight, and basically in every stage the weight multiplicity must be computed. But in our case the representations are multiplicity free, as revealed by simple counting, thus we are saved from this computationally complex task.

The weight diagram of $8_+$

$$\begin{matrix} &[0,0,1,0] & \\ & \downarrow \alpha_3& \\ & [0,1,-1,0] & \\ & \downarrow \alpha_2& \\ & [1,-1,0,1] & \\ & \alpha_1\swarrow \searrow\alpha_4 & \\ [-1,0,0,1] & & [1,0,0,-1] \\ &\alpha_4\searrow \swarrow \alpha_1 & \\ & [-1, 1 ,0, -1] & \\ & \downarrow \alpha_2 & \\ &[0, -1 ,1, 0] & \\ & \downarrow \alpha_3& \\ & [0, 0, -1, 0] & \end{matrix}$$

The weight diagram of $8_-$

\begin{matrix} &[0,0,0,1] & \\ & \downarrow \alpha_4& \\ & [0,1,0,-1] & \\ & \downarrow \alpha_2& \\ & [1,-1,1,0] & \\ & \alpha_1\swarrow \searrow\alpha_3 & \\ [-1,0,1,0] & & [1,0,-1,0] \\ &\alpha_3\searrow \swarrow \alpha_1 & \\ & [-1, 1 ,-1, 0] & \\ & \downarrow \alpha_2 & \\ &[0, -1 ,0, 1] & \\ & \downarrow \alpha_4& \\ & [0, 0, 0, -1] & \end{matrix}

Now the weights of the tensor product are the $64$ combinations of all possible sums of one weight from $8_+$ and one weight from $8_-$. These combinations are listed in the appendix at the end of this answer. The positive weights in this list are the candidates of the highest weights of the representation direct sum decomposition. One notices that this list includes the following positive weights:

$ [0, 0, 1, 1]$ : One copy

$ [1, 0, 0, 0]$ : 4 copies

The first weight is the highest weight of $56_v$, and the second one of $8_v$. But looking at the weight diagram of $8_v$:

\begin{matrix} &[1,0,0,0] & \\ & \downarrow \alpha_1& \\ & [-1,1,0,0] & \\ & \downarrow \alpha_2& \\ & [0,-1,1,1] & \\ & \alpha_3\swarrow \searrow\alpha_4 & \\ [0,0,-1,1] & & [0,0,1,-1] \\ &\alpha_4\searrow \swarrow \alpha_3 & \\ & [0, 1 ,-1, -1] & \\ & \downarrow \alpha_2 & \\ &[1, -1 ,0, 0] & \\ & \downarrow \alpha_1& \\ & [-1, 0, 0, 0] & \end{matrix}

one observes that the diagram is symmetric, the negative of any weight is also a weight, remembering that $56_v$ is the antisymmetric tensor product of 3 copies of $8_v$, thus every weight of $56_v$ is the antisymmetric tensor product of three distinct weights of $8_v$, we see that the highest weight can combine in three combinations to a distinct weight and its negative, thus the weight $[1, 0, 0, 0]$ will appear as an intermediate weight three times in $56_v$, thus three out of the four appearances of $[1, 0, 0, 0]$ in the tensor product are not highest weights, thus we are left with $56_v: [0, 0, 1, 1]$ and a single copy of $8_v: [1, 0, 0, 0]$. Of course the direct sum dimension agrees with the direct product dimension.

Appendix: The weights of the direct product representation

$$ \begin{matrix} &[ 0, 0, 1, 1]& \\ &[ 0, 1, -1, 1]& \\ &[ 1, -1, 0, 2]& \\ &[-1, 0, 0, 2]& \\ &[ 1, 0, 0, 0]& \\ &[-1, 1, 0, 0]& \\ &[ 0, -1, 1, 1]& \\ &[ 0, 0, -1, 1]& \\ &[ 0, 1, 1, -1]& \\ &[ 0, 2, -1, -1]& \\ &[ 1, 0, 0, 0]& \\ &[-1, 1, 0, 0]& \\ &[ 1, 1, 0, -2]& \\ &[-1, 2, 0, -2]& \\ &[ 0, 0, 1, -1]& \\ &[ 0, 1, -1, -1]& \\ &[ 1, -1, 2, 0]& \\ &[ 1, 0, 0, 0]& \\ &[ 2, -2, 1, 1]& \\ &[ 0, -1, 1, 1]& \\ &[ 2, -1, 1, -1]& \\ &[ 0, 0, 1, -1]& \\ &[ 1, -2, 2, 0]& \\ &[ 1, -1, 0, 0]& \\ &[-1, 0, 2, 0]& \\ &[-1, 1, 0, 0]& \\ &[ 0, -1, 1, 1]& \\ &[-2, 0, 1, 1]& \\ &[ 0, 0, 1, -1]& \\ &[-2, 1, 1, -1]& \\ &[-1, -1, 2, 0]& \\ &[-1, 0, 0, 0]& \\ &[ 1, 0, 0, 0]& \\ &[ 1, 1, -2, 0]& \\ &[ 2, -1, -1, 1]& \\ &[ 0, 0, -1, 1]& \\ &[ 2, 0, -1, -1]& \\ &[ 0, 1, -1, -1]& \\ &[ 1, -1, 0, 0]& \\ &[ 1, 0, -2, 0]& \\ &[-1, 1, 0, 0]& \\ &[-1, 2, -2, 0]& \\ &[ 0, 0, -1, 1]& \\ &[-2, 1, -1, 1]& \\ &[ 0, 1, -1, -1]& \\ &[-2, 2, -1, -1]& \\ &[-1, 0, 0, 0]& \\ &[-1, 1, -2, 0]& \\ &[ 0, -1, 1, 1]& \\ &[ 0, 0, -1, 1]& \\ &[ 1, -2, 0, 2]& \\ &[-1, -1, 0, 2]& \\ &[ 1, -1, 0, 0]& \\ &[-1, 0, 0, 0]& \\ &[ 0, -2, 1, 1]& \\ &[ 0, -1, -1, 1]& \\ &[ 0, 0, 1, -1]& \\ &[ 0, 1, -1, -1]& \\ &[ 1, -1, 0, 0]& \\ &[-1, 0, 0, 0]& \\ &[ 1, 0, 0, -2]& \\ &[-1, 1, 0, -2]& \\ &[ 0, -1, 1, -1]& \\ &[ 0, 0, -1, -1]& \end{matrix}$$

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If thats not very cumbersome, I'd hate to see a cumbersome calculation! –  BebopButUnsteady Jul 1 '13 at 20:19

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