1
$\begingroup$

I need to work out the weight systems for the fundamental representation $\mathbf{5}$ and the conjugate representation $\overline{\mathbf{5}}$. I'm not clear what this means. The $\mathbf{5}$ representation is of course just the representation of $SU\left(5\right)$ by itself. After picking a Cartan subalgebra as the diagonal matrices with zero trace, we can of course see that the roots are $L_i-L_j$ where $L_i$ picks out the $i^{th}$ element on the diagonal, and the weights are simply $L_i$ in this case.

It is supposed to be the case that I can use the weight systems of representations to show for instance that $\mathbf{5}\otimes \mathbf{5}=\mathbf{10}\oplus \mathbf{15}$.

$\endgroup$
0
$\begingroup$

The weights of an irrep $\lambda^*$ conjugate to $\lambda$ will be the negative of the weights in $\lambda$. The simplest way to handwave your way to this is to consider the (diagonal) transformations $$ \exp\left[i\sum_k \theta_k \Lambda_k\right]\, , \tag{1} $$ where $\Lambda_k$ is a Cartan element. Act on a state of definite weight $(w_1,\ldots,w_k)$ to produce $$ \exp\left[i\sum_k \theta_k \Lambda_k\right]\vert w_1\ldots w_k\rangle= \exp\left[i\sum_k \theta_k w_k\right]\vert w_1,\ldots w_k\rangle. $$ Now take the complex conjugate of (1) , which clearly produces $$ \exp\left[i\sum_k \theta_k q_k\right] \vert w_1,\ldots w_k\rangle^* $$ i.e. the weights are all reversed.

This is not a very formal proof but rather an intuitive justification of the rule given about as to the relation between the weights in $\lambda^*$ and $\lambda$.

FYI the simplest way to couple copies of the fundamental irrep is to use Young tableaux. In your specific case the result must be in the $SU(5)$ irreps labelled by the partitions $\{2\}$ and $\{1,1\}$, corresponding to Dynkin labels $(2,0,0,0)$ and $(0,1,0,0)$. The $(2000)$ is fully symmetric while the $(0100)$ is fully antisymmetric.

$\endgroup$

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.