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Supermultiplet dimensions from Young Tableaus

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In John Terning's book, on pages 14 and 15, there are lists of $\mathcal{N} = 2$ and $\mathcal{N} = 4$ supermultiplets, labeled in terms of the dimensions of the corresponding R-symmetry $d_R$ and spin-symmetry $2j+1$. I want to figure out a way to get all these numbers by Young Tableaus in a systematic way.

Of course, the $\mathcal{N}=2$ case is relatively straightforward. Its clear that the numbers in the labels for $\mathcal{N}=4$ can separately be recovered using

$4_{R} \otimes 4_{R} = 10_R \oplus 6_R$

for $SU(4)_R$ and tensor products of this basic identity with $4_R$. Of course, one can also do the same thing for the $SU(2)$ spin symmetry

$2_{SU(2)} \otimes 2_{SU(2)} = 3_{SU(2)} + 1_{SU(2)}$

But if one writes $(\textbf{R}, 2j+1)$ as the label, then how does one justify

$(\textbf{4}_R, 2)\otimes(\textbf{4}_R, 2) = (\textbf{10}_R,1) \oplus (\textbf{6}_R,3)$

or

$(\textbf{4}_R, 2)\otimes((\textbf{10}_R,1) \oplus (\textbf{6}_R,3)) = (\bar{\textbf{20}}, 2) + (\bar{\textbf{4}},4)$

What I'm asking is: how do you get this particular grouping?

This post imported from StackExchange Physics at 2015-06-02 11:45 (UTC), posted by SE-user leastaction
asked Jun 2, 2015 in Theoretical Physics by leastaction (425 points) [ no revision ]

1 Answer

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The way to justify this is to realize that when you write the label as  \((\textbf{R}, 2j+1)\) you are embedding the direct product of those groups into a larger group i.e.  \(SU(N) \otimes SU(M) \in SU(NM)\). Where the tensor indices of the larger group can be thought of as an ordered pair of indices \((a,\dot\alpha)\), where \(a=1,...,N\) and \(\dot\alpha = 1,...,M\) for this case we have N=4 and M=2.

In Terning's book we can see that each charge carries an order pair (which is embedded in the larger SU(NM) group) and has to be completely antisymmetric which every other charge. So if we have n charges we want to represent that as a completely antisymmetric rank n tensor living in the SU(NM) group. The way to represent this with Young's Tableaus is to think of it as n boxes arranged vertically. The trick is the embedded group has to transform the same way as the representation in the larger group when permutating the ordered indices.

For example. \((\textbf{4}_R, 2)\otimes(\textbf{4}_R, 2) = (\textbf{10}_R,1) \oplus (\textbf{6}_R,3)\)

This represents a completely antisymmetric rank 2 tensor in SU(8). The first thing to do when trying to come up with possible representations is to write down all the ways 2 (note this number is the same as rank of the embedding group) boxes can be arranged in SU(4) and SU(2), which happens to be the same 2 horizontal and 2 vertical or the 10 and 6 for SU(4) and the 3 and 1 for SU(2) respectively. (I don't know how to draw boxes here...). Where the boxes arranged horizontally are completely symmetric and the ones arranged vertically are completely antisymmetric. Since we need the permutation of ordered pair to be antisymmetric, the only way to do this is to have the symmetric rep in one group be ordered with the antisymmetric of the other rep, for permutations of the symmetric rep give you a plus while the antisymmetric give you a minus resulting in an overall minus.

The next rep is a little more complicated. \((\textbf{4}_R, 2)\otimes(\textbf{4}_R, 2)\otimes(\textbf{4}_R, 2) = (\bar{\textbf{20}}, 2) \oplus (\bar{\textbf{4}},4)\)

Here we have a rank 3 tensor. The way to arrange 3 boxes in SU(4) result in \(\bar 4, \bar{20''}, \bar{20} \) check out his appendix B.4 to align the box notation with those numbers. While the SU(2) only gives the 2 and 4 rep (note the 2 rep is drawn with 3 boxes the same way as the \bar{20}!!) The \((\bar{\textbf{4}},4)\) follows the same logic as previous, the \bar{4} is completely antisymmetric while the 4 is completely symmetric. The \((\bar{\textbf{20}}, 2)\) is more complicated, and I wasn't able to find a completely generic way to show it, but essentially when you take the tensor product of the two reps in the permutation group (S_n, n being the 3 in this case) you find that they can decomposed into a completely antisymmetric tensor (and actually a completely symmetric  tensor).

Georgi explains all of this in gory detail this in his book sections 15.2 and 1.21-24.

Long story short the decomposition of the representations should transform under permutations the same way as the rank n tensor embedded in the higher group. The left hand side is completely antisymmetric so too should the right hand side.

(Sorry the previous edits were made under some errors and I didn't know how to delete so I just redid it)

answered Jun 15, 2015 by Peter Anderson (205 points) [ revision history ]
edited Jun 19, 2015 by Peter Anderson

Thank you for the detailed reply @PeterAnderson! As it turns out, there is a more heuristic method which I managed to figure out, which is purely tableau-based. It is described in the comments of the original post on Physics.SE, from which this post was derived.

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