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?
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$\begingroup$ 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. $\endgroup$– user178876Nov 28, 2018 at 4:21
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$\begingroup$ Is this done simply by matrix multiplication of the Dynkin label? $\endgroup$– Joshua TilleyNov 28, 2018 at 4:33
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$\begingroup$ Yes, too long for a comment so I wrote an answer. $\endgroup$– user178876Nov 28, 2018 at 4:34
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$\begingroup$ Would Mathematics be a better home for this question? $\endgroup$– Qmechanic ♦Nov 28, 2018 at 5:17
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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.
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$\begingroup$ How did you get the weights' Dynkin labels? Also, shouldn't there be four weights, since the rank is four? $\endgroup$ Nov 28, 2018 at 10:33
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