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I wrote a Mathematica paclet that can be used to find irreducible representations of $SU(n)$. Specifically, given a product of $SU(n)$ multiplets, it can compute the corresponding sum. To test the paclet, I compare its output with various examples from books. It fails on two occasions, when I compare against Table 4.13.1 on page 83 in The Lie Algebras of su(N) by Walter Pfeifer. I also did the calculations by hand, but I obtain the same answer as my program.

My calculation: $$6 \otimes 8 = \color{red}{\bar{3}} \oplus \bar{15} \oplus 6 \oplus 24$$ Book: $$6 \otimes 8 = \color{red}{3} \oplus \bar{15} \oplus 6 \oplus 24$$

My calculation: $$15 \otimes \bar{15} = 1 \oplus \bar{10} \oplus \color{red}{10} \oplus 8 \oplus 8 \oplus \bar{35} \oplus 27 \oplus 27 \oplus 64 \oplus 35$$ Book: $$15 \otimes \bar{15} = 1 \oplus \bar{10} \oplus \color{red}{\bar{10}} \oplus 8 \oplus 8 \oplus \bar{35} \oplus 27 \oplus 27 \oplus 64 \oplus 35$$

Is this my mistake, or are these typos in the book?

Here is the explicit calculation of the first example, using Young tableaux: enter image description here

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    $\begingroup$ Hopefully someone more familiar with the physics ways of writing these things can help. But if you want this is "math" language, you are looking at tensor products (not direct products) of representations (not sure why they would be called multiplets). And for $SU(3)$ they would usually be indexed by two numbers, not one (by indexing them by their highest weight). $\endgroup$ Feb 22 '18 at 20:35
  • $\begingroup$ Ok, thanks. In physics it is customary to label the SU(3) multiplets by their dimension. So $3$ corresponds to $(1,0)$ and $\bar{3}$ corresponds to $(0,1)$. $\endgroup$
    – JEM_Mosig
    Feb 22 '18 at 20:53
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    $\begingroup$ The book is wrong also in the decomposition of $15\otimes \overline{15}$. Surely the decomposition must be its own dual. Meaning that $10$ and $\overline{10}$ must appear with the same multiplicity (both will appear once as per your calculation). $\endgroup$ Feb 22 '18 at 22:07
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    $\begingroup$ @JyrkiLahtonen Good observation. Yet another good reason to not label everything by dimensions, since that requires remembering which of those are self-dual. $\endgroup$ Feb 23 '18 at 7:05
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Following the recipe for the multiplicities of components of a tensor product outline by Jim Humphreys here. IIRC this recipe is a consequence of Weyl's character formula. See Humphreys' book for a proof. In my edition this is Exercise 8 in section 24.4 (Steinberg's formula)

Assuming that the fundamental dominant weights are indexed in such a way that the representation you denote by $6$ is $V(2\lambda_1)$. In the case of the adjoint representation there is no ambiguity and the 8-dimensional representation is the one denoted by $V(\lambda_1+\lambda_2)$.

The module $V(2\lambda_1)$ has six distinct weights, each with multiplicity one, namele $2\lambda_1, \lambda_2, -2\lambda_1+2\lambda_2, \lambda_1-\lambda_2,-\lambda_1$ and $-2\lambda_2$. Therefore in the tensor product we get summands with highest weights (add $\lambda_1+\lambda_2$ to the weights listed above): $3\lambda_1+\lambda_2$,$\lambda_1+2\lambda_2$, $-\lambda_1+3\lambda_2$, $2\lambda_1$, $\lambda_2$ and $\lambda_1-\lambda_2$. Among these six weight the non-dominant ones have a non-trivial stabilizer under the dot action, so we simply throw those out. The result is that $$ V(2\lambda_1)\otimes V(\lambda_1+\lambda_2)\simeq V(3\lambda_1+\lambda_2)\oplus V(\lambda_1+2\lambda_2)\oplus V(2\lambda_1)\oplus V(\lambda_2). $$ So the same six-dimensional rep $V(2\lambda_1)$ we used as a factor in the tensor product also appears as a summand. The module $V(\lambda_2)$ is the dual of the 3-dimensional module $V(\lambda_1)$, and I'm fairly sure that physicists denote it $\overline{3}$. This is good news in the sense that it says that you're right and your other source is wrong!

Of the other two summands Weyl's dimension formula says that $$ \dim V(3\lambda_1+\lambda_2)=\frac{4\cdot6\cdot2}{1\cdot2\cdot1}=24, $$ and $$ \dim V(\lambda_1+2\lambda_2)=\frac{2\cdot5\cdot3}{1\cdot2\cdot1}=15. $$ The latter module is the dual of $V(2\lambda_1+\lambda_2)$, obviously also 15-dimensional. I may be wrong, but IIRC the way physicists' notation plays out is that of the non-self-dual modules of $SU(3)$ the one whose highest weight $m_1\lambda_1+m_2\lambda_2$ satisfies the inequality $m_1>m_2$ is denoted without the overline, while the other (with highest weight $m_2\lambda_1+m_1\lambda_2$) gets the "bar".

Assuming this recollection is in line with the physicist's notation, we get, in the end $$ 6\otimes 8\simeq 24\oplus6\oplus\overline{15}\oplus\overline{3}. $$


Another argument leading to the same conclusion is that the weights of $3=V(\lambda_1)$ are in a different coset of the root lattice from the rest of the weights appearing in this decomposition. Therefore $\overline{3}=V(\lambda_2)$ can appear (and actually does appear) as a composition factor, but $3$ cannot.

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