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This post is an exact copy of one that I posted in Math's site. I do this copy because people there suggested me to do it since, apparentely, in Mathematics and Physics we use different conventions for this kind of topics and none there could understand mine. (You can see first post here: https://math.stackexchange.com/q/3134077/)

Imagine you have the Young tableu and the Dynkin numbers, $(q_1, q_2, ..., q_r)$, of the Lie algebra of $SU(n)$ which has $r$ simple roots. The way I assign Dynkin numbers is increasing its value from left to right so the $k$-th Dynkin number is the number of columns with $k$ boxes: $q_k$ columns made of $k$ boxes.

The Young tableau is the 'usual' one with columns that decreases in boxes from left to right. The calculation of the dimension gives you some number $d$ that is given by

$$d = \frac{N}{H}$$

Where $N$ is the product of the following numbers: in the highest left box for $SU(n)$ write an $n$ and going to the right, increase this number in one unit box per box. Going down, decrease the number in the same amount box per box. $N$ is the product of all those numbers. $H$ is the product of the hook numbers: in each box write the number of boxes that you cut going from right (out of tableaux) to left till you reach that box and then keep cutting boxes going down from that box. Do this for each box and the product of these numbers (hook numbers) is $H$.

Now, my question is: how can I know if this Young tableau corresponds to the representation $d$ or to the complex conjugated $\bar{d}$ since both of them have the same dimension?


My source is: http://www.th.physik.uni-bonn.de/nilles/people/luedeling/grouptheory/data/grouptheorynotes.pdf sections 6.6.1 and 6.11

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  • $\begingroup$ ? Before jumping into generality, for SU(3), compare and contrast (1,2) with (2,1), so 15 and its conjugate. What's your question here? Names? $\endgroup$ – Cosmas Zachos Mar 23 at 18:03
  • $\begingroup$ @CosmasZachos Yeah, but my problem is how you know (1, 2) is '15' and (2, 1) '$\bar{15}$'. Is this just a convention that do not affect physics? For instance in quark model '3' is associated to quarks while '$\bar{3}$' to antiquarks $\endgroup$ – Vicky Mar 24 at 21:04
  • $\begingroup$ I would not worry about it... try context. If there is a convention, I am not familiar with it. I would call a rep unbarred if it has more quarks, height-1 columns, than antiquarks, and vice versa for barred. $\endgroup$ – Cosmas Zachos Mar 24 at 21:26
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    $\begingroup$ What would you do if you had three different reps with the same dimension, a fairly common ocurence? (Look at the three 20 d reps of SU(4)!) Then, clearly, the capricious bar notation is unhelpful. $\endgroup$ – Cosmas Zachos Mar 26 at 16:38
  • $\begingroup$ @CosmasZachos Thanks for keeping this topic still alive! Three 20 d representations in SU(4)... good call. I've been thinking about the physical implications of all this things (conjugated, non-conjugated, now this new info, etc), and I have realised that what is physically relevant are the Young projectors that you can deduce from standard Young tableaux. I think this is the physically relevant object that is extracted from the Young tableaux associated with each rep since it projects into irreducible & invariant subspaces giving the physical states (cont.) $\endgroup$ – Vicky Mar 27 at 0:33
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In general there is no 1-to-1 map from dimension to representation, unlike $\mathrm{SU}(2)$ where you can label representations by their dimensions. That means that you can certainly have different Young tableaux with the same dimension.

The complex conjugate representation is obtained by "flipping" the Young tableau upside down and left to right and completing it to a rectangle with $N$ rows. The boxes that you have to add form the Young tableau of the conjugate representation. It is then conventional to say which is the "not conjugate" and which is the "conjugate".

As a consequence, you can immediately tell whether a representation is real or not: just look if the Young tableau is symmetric with respect to the diagonal in the NE direction. This is my best shot at making some visuals. The Young tableau on the left is flipped as explained and its complex conjugate is made up of the boxes colored in black on the right. This is for $\mathrm{SU}(4)$. $$ (1,2,2)=\begin{matrix} \square & \square & \square & \square & \square \\ \square & \square & \square & \square \\ \square & \square \\ \end{matrix} \;\underset{\text{flip}}{\longrightarrow}\; \begin{matrix} \blacksquare & \blacksquare &\blacksquare &\blacksquare &\blacksquare\\ \blacksquare&\blacksquare&\blacksquare&\square & \square \\ \blacksquare&\square & \square & \square & \square \\ \square & \square & \square & \square & \square \\ \end{matrix} = (2,2,1) $$

In the language of tensors doing this "flipping" and completing to a rectangle is the same as contracting by the invariant tensor $\epsilon_{i_1\ldots i_N}$ or $\epsilon^{i_1\ldots i_N}$. Let me be more precise: if the fundamental representation is a vector $q_i$, then the antifundamental is an antisymmetric tensor $$ q_{i_1\ldots i_{N-1}} \equiv \epsilon_{i_1\ldots i_{N-1} i_N}\bar{q}^{i_N}\,, $$ which is clearly equivalent to a vector with the upper index thanks to the $\epsilon$ tensor. In general any group of antisymmetrized indices can be contracted with the identity defined in this way $$ A_{\ldots [i_1 \ldots i_n]\ldots} = \frac{1}{(N-n)!}\epsilon_{i_1\ldots i_n j_{n+1}\ldots j_{N}}\,\epsilon^{k_1\ldots k_n j_{n+1}\ldots j_N}\,A_{\ldots [k_1 \ldots k_n]\ldots} \,. $$ This contraction doesn't lose any components in the tensor, so I can drop the fist $\epsilon$ and thus obtain a new tensor with $N-n$ indices antisymmetrized, rather then $n$.

As an example, take the symmetric product of $3$ fundamentals in $\mathrm{SU}(N)$. This will be a single row of length $3$, and thus will be completed to $N-1$ rows of length $3$. With the trick explained above one has $$ q_{(i j k)} \;\longrightarrow\; \epsilon_{i i_1\ldots i_{N-1}}\epsilon_{j j_1\ldots j_{N-1}}\epsilon_{k k_1\ldots k_{N-1}}\bar{q}^{(ijk)}\,. $$ So it has the structures of a Young tableaux with $N-1$ rows of length $3$ and is clearly the complex conjugate of the irrep we had before.

And if you have say an antisymmetric product of two fundamentals, contracting with the $\epsilon$ will give an antisymmetric product of $2$ antifundamentals $$ q_{[ij]}\;\longrightarrow\; \epsilon_{ij k_1\ldots k_{N-2}}\bar{q}^{[ij]}\,. $$

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    $\begingroup$ Why choose (1,2,2) to be “the representation” and (2,2,1) its conjugate rather than the other way around? Is this not the question? (Actually your tableaux are very well done!) $\endgroup$ – ZeroTheHero Mar 24 at 16:39
  • $\begingroup$ I don't think people have an obvious candidate for which one is the "conjugate" one. Normally one simply regards two representations to be conjugate to each other. Maybe you can just define the "not conugate" one to be the one with the least boxes. (I think the number of boxes always changes under conjugation unless the representation is real but 'm not sure) $\endgroup$ – MannyC Mar 24 at 16:50
  • $\begingroup$ Ok so we agree. and Yes the number of boxes for an irrep and its conjugate is different. I can't find a pattern in Slansky's table an my guess is there is no "rule" to declare an irrep conjugate to another: (5,0) is conjugate to (0,5) in Slansky's table but (5,0) has fewer boxes... $\endgroup$ – ZeroTheHero Mar 24 at 16:53
  • $\begingroup$ As a side note: I stole the idea on how to make the tableaux from another post in this question, which was canceled shortly after. So I can't take credit for it ^^' $\endgroup$ – MannyC Mar 24 at 16:54
  • $\begingroup$ @Mane.andrea So from your answers and comments, we still don't know why (1,2,2) is 'the representation' and (2,2,1) 'the conjugated', true? My problem is that mathematically maybe is just convention which is which, but physically the conjugated is, for instance, in quark model associated to antiquarks while the 'not conjugated' to quarks. $\endgroup$ – Vicky Mar 24 at 21:00
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There doesn't seem to be much of a convention, as you can see from the following table taken from

Slansky, Richard. "Group theory for unified model building." Physics reports 79.1 (1981): 1-128.

enter image description here This is a list of representations, with they Dynkin label, with the understand that the conjugate one is obtained by reversing the ordering of the Dynkin label. Slansky also provides table of "representations" for SU(4) etc, with no obvious pattern.

Note also that the representation conjugate to $\lambda$ is NOT obtained by conjugating the tableau. For instance, the $(1,0)$ of $\mathfrak{su}(3)$ corresponds to the partition $\{1\}$ but its conjugate $(0,1)$ corresponds to the partition $\{1,1\}$, in accordance with the tensor product decomposition $(1,0)\otimes (1,0)=(2,0)\oplus (0,1)$.

On a side note, for a partition $\{\lambda_1,\lambda_2,\lambda_3,\ldots,\lambda_q\}$ the corresponding Dunkin labels are $(\lambda_1-\lambda_2,\lambda_2-\lambda_3,\ldots)$ so that, corresponding to the partition $\{1,1\}$ the Dynkin label is $(0,1,0\ldots)$. Thus, $\lambda_1=p_1+p_2+p_3\ldots+p_q$, $\lambda_2=p_2+p_3+\ldots$, $\lambda_k=\sum_{i=k}^q p_i$.

Moreover, the weights in $\lambda^*$ are the negatives of those in $\lambda$.

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  • $\begingroup$ OP is right about the Dynkin label notation, and your discussion agrees with him/her. $p_k$ (called $q_k$ in the post) is the number of columns that have height $k$, it is therefore the difference of the length of two consecutive rows $\lambda_k - \lambda_{k+1}$. Also, if you conjugate the tableau of $(1,0) = \{1\}$ following the definition in my answer, you do get $(0,1) = \{1,1\}$. Is there a mismatch in our definitions? $\endgroup$ – MannyC Mar 24 at 16:01
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    $\begingroup$ @Mane.andrea thanks for your observation. Re-reading you are right and the definition of Dynkin labels is right but I had not seen it in this way. I will edit my post to remove that part. $\endgroup$ – ZeroTheHero Mar 24 at 16:32
  • $\begingroup$ Even though OP's notation was not wrong, that part that you erased in your post was correct too and it could have been useful for people more used to the other notation. Maybe it's worth it to still keep it as a side note? $\endgroup$ – MannyC Mar 24 at 16:53
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    $\begingroup$ @Mane.andrea did just that thanks for the feedback. $\endgroup$ – ZeroTheHero Mar 24 at 16:55

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