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For concreteness let's consider $SO(4)$.

The quantum numbers for the four states in the fundamental representations are (schematically)

$$ (1, 1) ,(-1, 1) ,(1, -1) ,(-1, -1 )$$

thus

$$ 4= \begin{pmatrix} (1, 1) \\(-1, 1) \\ (1, -1) \\ (-1, -1 ) \end{pmatrix} $$

If I want to compute $ 4 \otimes 4 = 1$ using tensors, I would compute naively

$$ 4 \otimes 4 = \begin{pmatrix} (1, 1) \\(-1, 1) \\ (1, -1) \\ (-1, -1 ) \end{pmatrix}^T \begin{pmatrix} (1, 1) \\(-1, 1) \\ (1, -1) \\ (-1, -1 ) \end{pmatrix} $$ $$=(1, 1) (1, 1) +(-1, 1)(-1, 1) +(1, -1) (1, -1) +(-1, -1 )(-1, -1 ) $$

which is not the one-dimensional representation of $SO(4)$. Correct would be

$$ 4 \otimes 4 = \begin{pmatrix} (1, 1) \\(-1, 1) \\ (1, -1) \\ (-1, -1 ) \end{pmatrix}^T \begin{pmatrix} (1, 1) \\(-1, 1) \\ (1, -1) \\ (-1, -1 ) \end{pmatrix} $$ $$=(-1, -1) (1, 1) +(1, -1)(-1, 1) +(-1,1) (1, -1) +(1, 1 )(-1, -1 ) $$

Why does transposing here change the quantum numbers?

For $SU(N)$ things are more transparent, because there we combine the conjugate fundamental with the fundamental to get a singlet. For $SU(3)$: $3 \otimes \bar 3= 1$.

Complex conjugation flips the quantum numbers and thus the usual scalar product of $3$ and $\bar 3$ yields the singlet.

Why and how does this happen for $SO(n)$?

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  • $\begingroup$ @Danu No we have $4 \otimes 4 = 1 + 15$, where $15$ is the adjoint of $SU(4)$. I'm just interested in the first part, i.e. the product of the 4s that yield a singlet and not an element of the adjoint $\endgroup$
    – jak
    Commented Jul 31, 2015 at 12:02
  • $\begingroup$ @Danu I know how to work with weights. What I'm trying to understand here are tensor methods. A basis vector for the fundamental $SU(4)$ rep is $\begin{pmatrix} 1 \\ 0 \\ 0 \\0 \end{pmatrix}$. From the discussion in the OP we know that this state has quantum numbers (1,1), which we could call $SO(4)$ color number 1. Equivalently the basis state $\begin{pmatrix} 0 \\ 1 \\ 0 \\0 \end{pmatrix}$ would have color number 2, which means quantum numbers (-1,1). $\endgroup$
    – jak
    Commented Jul 31, 2015 at 12:13
  • $\begingroup$ @Danu The naive scalar product of some fields in the 4-rep $\Psi$ carrying color number 2, which we would write as a a quadruplet $\begin{pmatrix} 0 \\ \Psi \\ 0 \\0 \end{pmatrix}$, does not yield something with $SO(4)$ quantum number zero, as it should be. For the $SU(n)$ group like $SU(3)$ this is no surprise since $3\times 3 \neq 1$. Instead we have $\bar 3 \times 3 =1$ and conjugating flips our quantum numbers such that we get zero quantum numbers from the naive scalar product as it should be. $\endgroup$
    – jak
    Commented Jul 31, 2015 at 12:15
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Danu
    Commented Jul 31, 2015 at 12:18

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