Why is ${\bf N}\otimes\bar{\bf N} \cong{\bf 1}\oplus\text{(the adjoint representation)}$?

Question

I just watched this lecture and there Susskind says that

$${\bf N}\otimes\bar{\bf N} ~\cong~{\bf 1}\oplus\text{(the adjoint representation)}$$ for the Lie group $G= SU(N)$. Unfortunately, he does not offer any explanation for this. Does anyone know some good explanation?

Answer 1

Any object in the fundamental representation has an index $i=1,\cdots,N$, i.e. $\Phi^i$ and transforms under $$ \delta_a \Phi^i \to i (T_a)^i{}_j \Phi^j $$ where $T$ is the generator of the fundamental representation. An adjoint field has index $a = 1,\cdots,N^2-1$, i.e. $\Phi^a$ and transforms as $$ \delta_a \Phi^b \to - f_{ac}{}^b \Phi^c $$ In particular, given any adjoint field, we can contract it with the generators to construct an $N \times N$ matrix as $$ \Phi^i{}_j \equiv \Phi^a (T_a)^i{}_j $$ This matrix is traceless, i.e. $\Phi^i{}_i = 0$ and therefore has a total of $N^2-1$ independent paramaters, which precisely matches the dimension of the adjoint rep. Now, what is the transformation of $\Phi^i{}_j$? This is $$ \delta_a \Phi^i{}_j \to - f_{ac}{}^b \Phi^c (T_b)^i{}_j = i (T_a)^i{}_k \Phi^k{}_j - i \Phi^i{}_k (T_a)^k{}_j $$ where in the last line, we used the Lie algebra $[T_a , T_b ] = i f_{ab}{}^c T_c$. Now, we use the fact that the generators are Hermitian, i.e. $T^T = {\bar T}$. Then we find the transformation of $\Phi^i{}_j$ is $$ \delta_a \Phi^i{}_j \to i (T_a)^i{}_k \Phi^k{}_j - i ({\bar T}_a)_j{}^k \Phi^i{}_k $$ Thus, we find that the first index of $\Phi$ transforms under the fundamental $N$ and the second index transforms under the anti-fundamental ${\bar N}$. Thus, $\Phi^i{}_j$ must transform under the $N \otimes {\bar N}$ representation.

On the other hand, the transformation of the trace part of $\Phi$ is $$ \delta_a \Phi^i{}_i \to i (T_a)^i{}_k \Phi^k{}_i - i \Phi^i{}_k (T_a)^k{}_i = 0 $$ Thus, the trace part transforms as a scalar. Putting this together, we find $$ \text{adjoint rep.} \oplus 1 = N \otimes {\bar N} $$

Answer 2

We have $\mathfrak{su}(n)\otimes\mathbb{C}\cong\mathfrak{sl}(n,\mathbb{C})$ acting on $N=\mathbb{C}^n$. Since $N\otimes N^\ast\cong\mathrm{End}(N)$ as representations, this is the representation of $\mathfrak{sl}(n,\mathbb{C})$ on $\mathfrak{gl}(n,\mathbb{C})$, which internally decomposes as

$$ \mathfrak{gl}(n,\mathbb{C})=\mathbb{C}\cdot\mathrm{Id}\oplus\mathfrak{sl}(n,\mathbb{C}). $$