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I know that in $SU(3)$ $$\mathbf{8}\otimes \mathbf{8} = \mathbf{27}+\mathbf{10}+\mathbf{\bar{10}}+\mathbf{8}+\mathbf{8}+\mathbf{1}. $$

How can one use this to compute $$\mathbf{10}\otimes \mathbf{8}\otimes \mathbf{8}\otimes \mathbf{8}?$$

Can one start with simplifying (dropping the bold notation)

$$\tag{1} \mathbf{10}\otimes \mathbf{8}\otimes \mathbf{8} = 10\otimes27 \\ +10\otimes10 \\+10\otimes\bar{10} \\+10\otimes8 \\+10\otimes8 \\+10\otimes1?$$

Is $(1)$ even ok?

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  • $\begingroup$ Could you give a little context, or at least definitions? To the best of my knowledge, 10 and 8 are integers, not three-by-three unitary matrices. $\endgroup$ – Neal Jan 24 '15 at 0:57
  • $\begingroup$ @Neal see the quark model $\endgroup$ – Omnomnomnom Jan 24 '15 at 1:21
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    $\begingroup$ Yes, tensor products of reps distribute over sums of reps, so (1) is ok. I'm not sure how, e.g. $10\otimes 27$ decomposes. $\endgroup$ – Jason DeVito Jan 24 '15 at 1:50
  • $\begingroup$ Can one reduce 10x1? I strongly suspect no,but I have to ask. $\endgroup$ – Your Majesty Jan 24 '15 at 1:56
  • $\begingroup$ @Omnomnomnom Thanks! $\endgroup$ – Neal Jan 24 '15 at 3:04
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Here is a portion of the answer. All of these calculations are coming from here.

In terms of that website, we think of $SU(3)$ as $A_2$. Then, the $10$ dim rep has heighest weight $(3,0)$ in their notation (and the $\overline{10}$ has heighest weight $(0,3)$). The $27$ d rep is $(2,2)$. The notation the website uses is $X[3,0]$, $X[0,3]$, or $X[2,2]$ respectively.

According to that website, we have

$$10\otimes 27 = 1X[5,2] +1X[3,3] +1X[4,1] +1X[1,4] +1X[2,2] +1X[3,0] +1X[0,3] +1X[1,1],$$ or in your notation,

$$10\otimes 27 = 81 + 64 + 35 + \overline{35} + 27 + 10 + \overline{10} + 8$$

A small disclaimer: in $SU(3)$, the dimension of a representation does not determine the representation, even discounting conjugate representations. For example, $X[2,1]$, $X[4,0]$, $X[1,2]$, and $X[0,4]$ are each distinct irreducible $15$ dimensional representations. So, notation like $8$, $10$, $\overline{10}$, while unambiguous, would be ambiguous applied to the number $15$.

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  • $\begingroup$ To answer your comment, yes, $10\otimes 1 = 10$. Also, I'm not sure what size inputs the program I linked to can handle (I think it can only handle single digit highest weights). $\endgroup$ – Jason DeVito Jan 24 '15 at 2:10
  • $\begingroup$ Great answer @Jason DeVito. Where did you find the dictionary for that website's notation? $\endgroup$ – Your Majesty Jan 26 '15 at 15:06
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    $\begingroup$ @LoveLearning: I wish I had a nice answer. I used the website quite a bit about 6-7 years ago when trying to figure out some stuff for my thesis, and I remember a lot of how it worked from then. The $A_2 \iff SU(3)$ notation is fairly standard (see classification of simple Lie algebras), and the recognition that, e.g., $X[3,0]\iff 10$ came from plugging things into their "dimension of a module" option, and noting that dimension $10$ only occurred for $X[3,0]$ and $X[0,3]$. $\endgroup$ – Jason DeVito Jan 26 '15 at 16:26
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    $\begingroup$ Well, I kind of cheated. I don't know your notation well enough to say which one is barred, so in my head, I defined $X[4,1] = 35$. In terms of the Dynkin indices, it then follows that $X[1,4] = \overline{35}$. (I'm actually heading to lunch, then class, so I won't be able to respond for another 4 hours or so...) $\endgroup$ – Jason DeVito Jan 26 '15 at 16:45
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    $\begingroup$ Not to my knowledge. The only notation I've ever used is by labeling the nodes in the Dynkin diagram by natural numbers, or variations on that. $\endgroup$ – Jason DeVito Jan 27 '15 at 14:09

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