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I'm following the rules in this document to combine irreps of $SU(N)$ using Young tableaux.

If I'm not mistaken the product of two irreps should be symmetrical, that is $A \otimes B = B \otimes A$. I'm interested in knowing if a product of irreps yields a trivial representation (a "singlet") in its decomposition or not.

I tried an example to get the hand of Young tableaux, that is the product $(0,1,1,0)\otimes(1,0,0,1)$ of $SU(5)$. In term of Young tableaux, that is (sorry I don't know what is the best way to draw Young tableaux here):

$$\begin{matrix} \square & \square \\ \square & \square \\ \square & \\ \end{matrix}\quad \otimes \quad \begin{matrix} \square & \square \\ \square & \\ \square & \\ \square & \\ \end{matrix}$$

Using the rules in the link above, I identify each row of the 2nd diagram with a single color (or letter usually). Then I place the boxes one by one on the first diagram and I make sure that each color doesn't appear more than once in the same column (because it is antisymmetrized).

Following this rule it seems I can build a singlet out of this product: $$\begin{matrix} \square & \square \\ \square & \square \\ \square & \\ \end{matrix}\quad \otimes \quad \begin{matrix} \color{red}\blacksquare & \color{red}\blacksquare \\ \color{blue}\blacksquare & \\ \color{green}\blacksquare & \\ \color{orange}\blacksquare & \\ \end{matrix} \quad \rightarrow \quad \begin{matrix} \square & \square \\ \square & \square \\ \square & \color{red}\blacksquare \\ \color{red}\blacksquare & \color{blue}\blacksquare \\ \color{orange}\blacksquare & \color{green}\blacksquare \end{matrix}\ \ = 1.$$

But somehow if I try the other way aroung, I can't "fill" the columns because I must put two identical colors (symmetric) into the same column (antisymmetric):

$$\begin{matrix} \square & \square \\ \square & \\ \square & \\ \square & \\ \end{matrix} \quad \otimes \quad \begin{matrix} \color{red}\blacksquare & \color{red}\blacksquare \\ \color{blue}\blacksquare & \color{blue} \blacksquare \\ \color{green}\blacksquare & \\ \end{matrix} \quad \rightarrow \quad \begin{matrix} \square & \square \\ \square & \color{red}\blacksquare \\ \square & \color{blue}\blacksquare \\ \square & \color{blue}\blacksquare \\ \color{red}\blacksquare & \color{green}\blacksquare \end{matrix}\ \ = 0.$$

Clearly there's something wrong with the first result. I think it must vanish, probably because some antisymmetric relation between the yellow and green box, but I haven't found any reference for this.

The only rule close to this would be rule #3 in the link above, which reads (adapted to the color code used here):

  1. Starting in the second row (where the [blue boxes] were initially) add the [blue boxes] subject to the constraint that, reading from right to left starting at the right end of the first row and then moving on to the second row, the number of [red boxes] must be ≥ the number of [blue boxes] (≥ the number of [green boxes] ,etc.) at each point in the reading process.

I don't see how my "calculation" above doesn't satisfy that rule. Thanks for helping me understand what's going on.

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I'm interested in knowing if a product of irreps yields a trivial representation (a "singlet") in its decomposition or not.

If you only need to know this then note that the trivial representations appears in the product $A\otimes B$ with multiplicity one if $A = \bar{B}$ and with multiplicity zero otherwise.

The conjugate representation $\bar{B}$ can be obtained as a "flipped" Young Tableau as explained in this answer.

The reason is that the multiplicity of $C$ in $A\otimes B$ is the same as the number of trivial representations in the triple product $A\otimes B \otimes \bar{C} = A\otimes \bar{C}\otimes{B}$, which in turn is the multiplicity of $\bar{B}$ in $A\otimes \bar{C}$. When $\bar{C}$ is the trivial representation this means the multiplicity of $\bar{B}$ in $A$, which is one if they are the same or zero otherwise.


See e.g. Di Francesco - Mathieu - Sénéchal, Conformal field theory, $\,§13.5$.

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  • $\begingroup$ Yes, that's what I was trying to show "visually", ie the only way to obtain a singlet is to "fill" the columns with the dual representation. I was surprised when I saw I could get a singlet from $(1,0,0,1)\otimes(0,1,1,0)$ even though they are not the dual of each other, which strongly hinted that I missed something. But thanks anyway for the explanation, I never saw it that way. $\endgroup$ – Jasmeru Apr 14 at 13:35
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I realized my mistake the minute I posted my answer... Anyway I'll leave it here since it can serve as a reference.

The problem with my first product is that we can't actually get this configuration:

$$ \begin{matrix} \square & \square \\ \square & \square \\ \square & \color{red}\blacksquare \\ \color{red}\blacksquare & \color{blue}\blacksquare \\ \color{orange}\blacksquare & \color{green}\blacksquare \end{matrix}\qquad \textrm{from} \qquad \begin{matrix} \color{red}\blacksquare & \color{red}\blacksquare \\ \color{blue}\blacksquare & \\ \color{green}\blacksquare & \\ \color{orange}\blacksquare & \\ \end{matrix} $$

since we must place the green box first and it would go in the lower left corner. However, this configuration:

$$ \begin{matrix} \square & \square \\ \square & \square \\ \square & \color{red}\blacksquare \\ \color{red}\blacksquare & \color{blue}\blacksquare \\ \color{green}\blacksquare & \color{orange}\blacksquare \end{matrix}$$

vanishes since it violates rule #3 I quoted in my question: counting from right to left and from the first row to the last, the tableau should always satisfy: #Red ≥ #Blue ≥ #Green ≥ #Yellow (the order is determined by the initial order I chose; we usually use letters as labels so that #$a$ ≥ #$b$ ≥ ... ).

In this case the two products seem to give the same answer (at least they both don't yield a singlet in their decomposition).

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