I have been studying Young Tableaux representation from youtube to represent $2\times 2$ and other examples to in $SU(n)$ symmetry. But i am unable to understand nor able to find relevant answers of how to represent conjugate i.e how can one write $\bar{3}$ in a square box in $SU(3)$ representation is actually 2 and $\bar{6}$ in $SU(3)$ is represented as 3 in $SU(3)$. By saying 3 i mean i sy its dimension.

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    $\begingroup$ Perhaps this answers the question? $\endgroup$
    – MannyC
    May 17, 2020 at 8:03
  • $\begingroup$ @MannyC i read the material but unable to grasp what he want to say.Can you explain just by starting of with simple examples as mentioned in my question and showing how the dimensions are calculated for conjugate $\endgroup$ May 17, 2020 at 12:55
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    $\begingroup$ WP. $\endgroup$ May 17, 2020 at 15:00
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    $\begingroup$ The conjugate rep of r is the Young tableau, flipped upside down, which you tack at the bottom of that of r to make all columns of length n in SU(n). Just try it, and put it in your question, if you were not comfortable with your answer. $\endgroup$ May 17, 2020 at 17:01
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    $\begingroup$ Does this answer your question? Complex conjugated representation and its Young tableaux $\endgroup$ May 20, 2020 at 14:42

1 Answer 1


The confusion here is that conjugating a Young diagram does conjugate the representation. This is compounded by the regrettable use of the dimension to label the representation.

In the specific case of $\mathfrak{su}(3)$ (and more generally for $\mathfrak{su}(n)$), a better solution is to use the Dynkin labels $(a,b)$.

For $\mathfrak{su}(3)$, an irrep with Dynkin label $(a,b)$ corresponds to a two-rowed Young diagram with $a+b$ boxes on the first row, and $b$ boxes on the second row: $\{a+b,b\}$. The advantage of the Dynkin labels is that the representation conjugate to $(a,b)$ is just $(b,a)$. It will have the same dimension as $(a,b)$ but does not correspond to the diagram conjugate $\{a+b,b\}$. Indeed it's not hard to see that the diagram conjugate to $\{a+b,b\}$ will have in $a+b$ rows, so it cannot describe an $\mathfrak{su}(3)$ irrep if $a+b>3$. The irrep $(b,a)$ is associated with the diagram $\{b+a,a\}$. Note that the irreps $(b,a)$ is associated with a Young diagram having a different number of boxes than its conjugate irrep $(a,b)$.

There are standard formulas to obtain the dimension of the representation in terms of Dynkin labels: for $\mathfrak{su}(3)$ we have \begin{align} \hbox{Dim}(a,b)=\frac{1}{2}(a+1)(a+b+2)(b+1) \tag{1} \end{align} so that, for the irrep corresponding to a single box, $(a,b)=(1,0)$, with Young diagram $\{1\}$ (a single box) and dimension $(2\times 3\times 1)/2=3$. This is what you call $\textbf{3}$.

The conjugate irrep is $(0,1)$, which you call $\bar{\textbf{3}}$, and is associated to Young diagram $\{1,1\}$, i.e. two boxes in a single column. Its dimension is also $3$ since the dimensionality formula (1) is symmetric under the interchange of $a$ and $b$.


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