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My question is about the reduction of a representation of a group $SU(5)$ to irreps of the subgroup $SU(3)\times SU(2) \times U(1)$.

For example the weights of the 10 dimensional representation of SU(5) are

enter image description here

One can identify the irreps of the subgroup by regrouping the dynkin labels into $((a_3 a_4) ,(a_1), a_2)$ such that (denoting $-1$ by $\bar{1}$): $$ (1,1)_{Y} \rightarrow \left\{ \begin{array}{l l} (0 0,0,1 ) \end{array} \right. $$

$$ (\overline{3},1)_{Y} \rightarrow \left\{ \begin{array}{l l} (0 1,(0),\bar{1}) \\ (1 \bar{1},(0),\bar{1})\\ (\bar{1}0,(0),0) \end{array} \right. $$

$$ (3,2)_{Y} \rightarrow \left\{ \begin{array}{l l} (1 0,1,\bar{1}) \\ (\bar{1} 1,\bar{1},1)\\ (0\bar{1},\bar{1},1)\\ (1 0,\bar{1},0)\\ (\bar{1}1,1,0)\\ (0\bar{1},1,0) \end{array} \right. $$

My problem is: how can I derive the $Y$ charge of the $U(1)$ factor for each of these from the Dynkin labels?


Edit

The metrictensor for SU(5) is thus

$$G= \frac{1}{5}\left( \begin{array}{cccc} 4 & 3 & 2 & 1 \\ 3 & 6 & 4 & 2 \\ 2 & 4 & 6 & 3 \\ 1 & 2 & 3 & 4 \end{array} \right). $$

However in the reference, Slansky, on page 84 the same exercise is done but the axis have negative values... $$\tilde{Y}^W = \frac{1}{3} [-2 \;1\, -1\; 2]. $$

How come they do not agree?

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Just a resource recommendation: see Weinberg's appendix in Classical Solutions in Quantum Field Theory. –  JamalS May 30 at 10:09

3 Answers 3

It is very simple using the young tableaux. Indeed, the $SU(M+N)$ decomposes in $SU(M)\times SU(N)\times U(1)$ where the hypercharge is identified (up to an overall normalization) with the (traceless) block diagonal matrix $diag(N,\ldots,N,-M,\ldots,-M)$ where the two blocks are each multiple of the identity with dimensions $M$ and $N$ respectively. Therefore, an irreducible representation of $SU(M)\times SU(N)$ given by two young tableaux with $m$ and $n$ boxes will have hypercharge $y=m N -nM$.

Example in $SU(3)\times SU(2)\times U(1)\subset SU(5)$: the adjoint 24 contains $(3,2)_{y=5}\in SU(3)\times SU(2)\times U(1)$ because the $3\in SU(3)$ comes from $n=4$ boxes whereas the $2\in SU(2)$ from $n=1$ boxes, and hence $y=4\cdot 2-1\cdot 3=5$.

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In this answer I'll be following Slansky's review: "Group theory for unified model building" and using the data from the review and the same notations.

The "U(1)" factors in the unbroken group correspond to "central charges" which must commute with the non-Abelian factors. It is not very hard to prove that their eigenvalues on the fundamental weights are given by corresponding component of the weight in the root basis. These components are given by:

$$ \bar{\lambda}_i = G_{ij} a_j$$

Where $a_j$ are the components in the Dynkin basis as given in the question. $G_{ij}$ is the metric tensor defined in terms of the Cartan matrix in equation (4.11) in Slansky. The values of this metric for the whole Cartan classification are given in table 7.page 82.

In the example given in the question when we identify the $SU(3)$ Dynkin labels with the first two labels of the $SU(5)$ weight and the $SU(2)$ label with the last one, then the (central) U(1) charge is the third component of the weight in the root basis.

Thus it is given by the scalar product of the third row of the metric tensor, which we can read as: $\frac{1}{5}[2, 4, 6, 3]$ with the weight.

The results of the scalar product is $\frac{4}{5}$ with the first $3$ weights, $\frac{-1}{5}$ with the next $6$ weights and $\frac{-6}{5}$ with the last weight.

Now, the branching does not impose any normalization requirements on the central charge. (There are outside conditions which may be used for that, but this will be outside the scope of the question). The normalization is imposed so the subrepresentation identified with the quarks will have the required charge namely $\frac{1}{3}$. This means that we must choose a normalization factor such that the charge $\frac{-1}{5}$ subspace becomes $\frac{1}{3}$. Thus the normalization factor is $\frac{-5}{3}$, therefore the corresponding charges must be $\frac{-4}{3}$ for the first $\frac{1}{3}$ for the next $6$ weights and $2$ for the singlet.

Of course each irreducible component in the decomposition is characterized by a single charge as should be.

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Thank you very much for this answer. However I'm somewhat trouble with the reference you gave: although they do exactly the same exercise they have a different axis: see above I have expanded the question. –  user41746 Jun 5 at 12:04
    
@user41746 $SU(5)$ is $A_4$, so the metric tensor should be $4\times 4$ matrix, please see the first matrix in table 7 page 82. Its third row for n=4 is just the given vector. I can’t see this exercise on page 84 (This page belongs to the appendix and it contains tables 10 and 11a. However, on page 16 equation (3.3), Slansky obtained the same hypercharges as in the question. –  David Bar Moshe Jun 5 at 12:30
    
Indeed, I had copied the wrong matrix (It has been corrected above). I think we are not referring to the same document... I referring to the example in Slanksy is in Chapter 6 equation (6.9) in cds.cern.ch/record/134739/files/…, in this document p 16 is about the irrep of left-handed fermions... –  user41746 Jun 5 at 13:20
    
@user41746, Sorry I noticed that I used $a_1, a_2$ for the $SU(3)$ labels and $a_4$ for the $SU(2)$ and in your example you used $a_3, a_4$ for $SU(3)$ labels and $a_1$ for SU(2). Thus you should use the second row in $G$ for your example. You will get the same result with the normalization factor $\frac{1}{3}$. –  David Bar Moshe Jun 5 at 14:20
    
@user41746, cont.Now, Slansky insists to identify the highest weight of the $SU(5) $ representation with the highest weight of the quark representation after the symmetry breaking. For that he needs to act on the weights with a “projection matrix” given in equation 6.7. After the action of the projection matrix the fourth component of the weights is lost, but nevertheless, the dual U(1) vector can be obtained by insisting that it has a constant scalar product with each subrepresentation weights. As you can see all the exercises lead to the same result for the hypercharges. –  David Bar Moshe Jun 5 at 14:21

In general, the $U(1)$ charges are not fixed uniquely from the representation. Still, you can find a linear combination of Dynkin labels that gives the $U(1)$ charge once you fix the charge of one single state. In the geometric picture, the $U(1)$ charges correspond to axes that you fix in weight space. This is unique if you know the quantum numbers of one state (e.g. $(1, 1)_Y$ should be the positron, of which you know the hypercharge, etc.).

Slansky deals with this in detail in his classic review "Group theory for unified model building", Chap. 6, around eq. (6.9).

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