3
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ Please include the relevant contents of the link into the post, in particular: 1. Define which group we are talking about. 2. Define $N$. 3. Clarify whether $\times$ is supposed to be the direct sum $\oplus$ or the tensor product $\otimes$, since it is mathematically the former, but physicists often seem to use $\times$ where they should use $\otimes$. $\endgroup$ Apr 14 '15 at 12:00
  • $\begingroup$ Related: physics.stackexchange.com/q/89173/2451 $\endgroup$
    – Qmechanic
    Apr 14 '15 at 12:24
  • $\begingroup$ There is absolutely no physics at all in this question. Why aren't people voting to move this to Math? $\endgroup$
    – DanielSank
    Apr 14 '15 at 16:43
1
$\begingroup$

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 $$ \text{adjoint rep.} \oplus 1 = N \otimes {\bar N} $$

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.