# Evaluating $n \otimes_A n^*$ in $SU(n)$

In "Quantum Field Theory in a Nutshell" pg424 the author (Zee) writes:

$$(n\oplus n^*)\otimes_A(n \oplus n^*)\quad\cong\quad(n^2-1)\oplus 1 \oplus n(n-1)/2 \oplus ((n(n-1))/2)^*$$

From what I understand to do the $n\otimes_A n$ we look at the contractions with the Levi-Civita tensor e.g. in SU(3) $$3\otimes_A 3=\varepsilon^{ijk}\psi_i\phi_j=3^*$$ but I don't understand how to do the same for $(n \otimes_A n^*)\oplus(n^* \otimes_A n)$ to get the adjoint (if it is actually the adjiont and not some anti-symmetric version?) and singlet reps. Please can someone explain?

Note: I have recently have being asking similar questions on MSE but apparently the notation used here is only common in physics - that along with its relations to the standard model I thought it was best to post this one here.

$${\bf n} \otimes_A {\bf n} \quad\cong\quad \begin{array}{c} [~~]\cr [~~] \end{array} \quad\cong\quad {\bf\frac{n(n-1)}{2}},\tag{1}$$ and $${\bf n} \otimes_A {\bf n}^{\ast} \quad\oplus\quad {\bf n}^{\ast} \otimes_A {\bf n} \quad\cong\quad {\bf n}^{\ast}\otimes {\bf n}$$ $$\quad\cong\quad \begin{array}{c} [~~]\cr [~~]\cr \vdots\cr [~~] \end{array} \otimes [~~] \quad\cong\quad \begin{array}{c} [~~]\cr [~~]\cr \vdots\cr [~~]\cr [~~] \end{array} \quad\oplus\quad \begin{array}{cc} [~~]&[~~]\cr [~~]\cr \vdots\cr [~~]\end{array} \quad\cong\quad {\bf 1} \quad\oplus\quad {\bf(n^2-1)},\tag{2}$$ etc. (The number of boxes in each term of eq. (2) is supposed to be $n$.)
• Sorry to be a pain. In my last comment (now deleted) I asked about the distributive law. What I was actually more concerned about is the first isomorphism in (2) between ${\bf n} \otimes_A {\bf n}^{\ast} \quad\oplus\quad {\bf n}^{\ast} \otimes_A {\bf n} \quad\cong\quad {\bf n}^{\ast}\otimes {\bf n}$ do you know of an explicit way to see that this holds? (Thanks again) – Quantum spaghettification Jan 8 '18 at 19:53
• OK I think I can answer my own comment. Let $(\psi_\mu \phi_\nu-\psi_\nu \phi_\mu, 0)\mapsto \psi_\mu \phi_\nu$ and $(0,\psi_\mu \phi_\nu-\psi_\nu \phi_\mu,)\mapsto -\psi_\mu \phi_\nu$. I think this will then be an isomorphism – Quantum spaghettification Jan 8 '18 at 19:58