# Branching of $SU\left(5\right)$

In the context of branching rules, what is a projection matrix for a subgroup. For instance, the projection matrix for the subgroup $$SU\left(2\right)\times SU\left(3\right)$$ of $$SU\left(5\right)$$ is apparently $$\left(\begin{array}{cccc} 0 & 1 & 1 & 0\\ 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 \end{array}\right)$$ What does this mean, and how do I use it?

• You can use it to get the Dynkin labels of the representations under the subgroup from the Dynkin labels of the representations of the supergroup.
– user178876
Nov 28, 2018 at 4:21
• Is this done simply by matrix multiplication of the Dynkin label? Nov 28, 2018 at 4:33
• Yes, too long for a comment so I wrote an answer.
– user178876
Nov 28, 2018 at 4:34
• Would Mathematics be a better home for this question? Nov 28, 2018 at 5:17

You can use the projection matrix to find out into which representations a given representation branches. In your example SU(5)$$\to$$SU(3)$$\times$$SU(2), the 5-plet has the weights with Dynkin labels $$(1, 0, 0, 0),\quad(-1, 1, 0, 0),\quad(0, -1, 1, 0),\quad(0, 0, -1, 1),\quad(0, 0, 0, -1)\;.$$ They get mapped under $$P$$, i.e. by multiplying them by the $$P$$ matrix, to $$(0| 1, 0),\quad (1| 0, 0),\quad (0| -1, 1),\quad (-1| 0, 0),\quad(0| 0, -1).$$ where I separated the SU(2) part from the SU(3) part by a bar |''. After reordering, this becomes $$\{ (1| 0, 0),(-1| 0, 0)\}\quad\text{and}\quad\{(0| 1, 0),(0| -1, 1),(0| 0, -1)\}\;,$$ i.e. an SU(2) doublet plus an SU(3) triplet, as expected.