In Srednicki's QFT book (see pg. 605) he writes that the fundamental representation of $SU(5)$ may be written as (97.2),

$$ 5\to \left(3,1,-\frac{1}{3}\right)\otimes \left(1,2, +\frac{1}{2}\right).\tag{84.12/97.2} $$

He then proceeds to state that $5\otimes 5$ transforms as (Eq. 97.4)

$$\begin{align} 5\otimes 5\to &\left(6,1, -\frac{2}{3}\right)_{\rm S} \oplus \left(3,2, +\frac{1}{6}\right)_{\rm S} \oplus \left(1,3, +1\right)_{\rm S}\\[0.25cm] &\oplus \left(\bar{3},1, -\frac{2}{3}\right)_{\rm A} \oplus \left(3,2, +\frac{1}{6}\right)_{\rm A} \oplus \left(1,1,+1\right)_{\rm A}. \end{align}\tag{97.4}$$

How can one derive this formula? In particular, what are the rules for computing the tensor products? How can you tell that a specific combination is symmetric or anti-symmetric?

I know for example, that one simply adds the hypercharge, but I am unsure how to deal with the representations of $SU(3)$ and $SU(2)$.

Any help or resources would be appreciated!

Edit: I should add that I am familiar with Young tableaux, and therefore taking tensor products for two representations of the same group. However, I am unsure how to extend this to tensor products of groups as in the above example.

  • $\begingroup$ Almost all standard textbooks on Lie algebra in particle physics discussed this thing. One can check Georgi or Zee's books. $\endgroup$
    – Mass
    Oct 8, 2021 at 1:07
  • 1
    $\begingroup$ You have written your $\boldsymbol{5}$ as a tensor product so you should clarify the notation. Most likely you need the branching rules from $\mathfrak{su}(5)$ to whatever other subalgebra you are using. $\endgroup$ Oct 8, 2021 at 1:12
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/313930/2451 $\endgroup$
    – Qmechanic
    Oct 8, 2021 at 12:25

1 Answer 1


One can start with the fundamental representation of $SU(5)$ as

\begin{equation} \mathbf{[5]} = \begin{pmatrix} q_{i}\\ \ell_{j} \end{pmatrix} = \begin{pmatrix} q_{1}\\ q_{2}\\ q_{3}\\ \ell_{1} \\ \ell_{3} \end{pmatrix} \end{equation}

where $q_{i}\,(i=1, 2, 3)$ and $\ell_{j}\,(j=1, 2)$ are the quark and lepton fields respectively.

Now, one can embed $SU(3)$ and $SU(2)$, in $SU(5)$ in the following way,

\begin{equation} SU(5) = \begin{pmatrix} SU(3)_{c} & \\ & SU(2)_{L} \end{pmatrix} \end{equation}

Then, following the transformation rule of fundamental representation $\mathbf{[5]}$, we can write \begin{equation*} \mathbf{[5]} \rightarrow \begin{pmatrix} SU(3)_{c} & \\ & SU(2)_{L} \\ \end{pmatrix} \begin{pmatrix} q_{i} \\ \ell_{r} \end{pmatrix} \end{equation*} Recall that quarks are triplet under $SU(3)_{c}$ and leptons are color blind, which simply put in the following way, \begin{eqnarray} \mathbf{[5]} & = & \mathbf{(3,1)}\oplus \mathbf{(1,2)} \\ \mathbf{[\bar{5}]} & = & \mathbf{(\bar{3},1)} \oplus \mathbf{(1,\bar{2})} \end{eqnarray} where $\mathbf{[\bar{5}]}$ is for anti-fundamental representation. Both up and down quark fields are triplets under the color group. One immediate question is, what fields corresponds to the $\mathbf{(\bar{3},1)}$ or $\mathbf{(3,1)}$?

Recall that one of the generators $SU(5)$ is electric charge $Q$ itself and which is diagonal! Therefore \begin{equation} \sum\mathrm{Eigenvalues} = 0 \end{equation}

which fixes $\mathbf{[\bar{3}]}$ in $\mathbf{[\bar{5}}]$ as down quarks. \begin{equation*} \bar{\mathbf{[5]}} = (\mathbf{3},\mathbf{1}) \oplus (\mathbf{1},\mathbf{2}) = \begin{pmatrix} d^{rc} & d^{bc} & d^{gc} & e^{-} & -\nu_e \end{pmatrix}_L \end{equation*} From this one can immediately check \begin{equation*} 3Q_{d^{c}}+Q_{e^{-}} = 0, \Longrightarrow Q_{d^{c}} = +\frac{1}{3}e \end{equation*} It says, fixing the charge of the electron automatically fixes the charge of the down-quarks. Which is indeed the quantization of electric charge. The remaining standard model matter fields can be contained in $[\mathbf{10}]$ representation. To achieve that we need to do some amount of group theory. Consider the following product \begin{align} \mathbf{[5]}\otimes\mathbf{[5]} &= [(\mathbf{3},\mathbf{1}_{2})\oplus (\mathbf{1}_{3},\mathbf{2})] \otimes [(\mathbf{3},\mathbf{1}_{2})\oplus (\mathbf{1}_{3},\mathbf{2})] \\ &= (\mathbf{3}\otimes\mathbf{3} ,\mathbf{1}_{2})\oplus (\mathbf{3},\mathbf{2})\oplus(\mathbf{3},\mathbf{2})\oplus (\mathbf{1_{3}},\mathbf{2}\otimes\mathbf{2})\\ &= (\mathbf{6},\mathbf{1}_{2})\oplus (\mathbf{\bar{3}},\mathbf{1}_{2})\oplus 2(\mathbf{3},\mathbf{2})\oplus (\mathbf{1_{3}},\mathbf{3})\oplus(\mathbf{1_{3}},\mathbf{1_{2}}) \end{align} where we have used tensor product decomposition of $SU(3)$ and $SU(2)$ Lie algebra, namely \begin{align} \mathbf{[3]} \otimes \mathbf{[3]} &= \mathbf{[6]}_{S}\oplus\mathbf{[\bar{3}}_{A}]\\ \mathbf{[2]} \otimes \mathbf{[2]} &= \mathbf{[3]}_{S}\oplus\mathbf{[1]}_{A} \end{align}

Now, from the general rule of decomposition for $SU(n)$

\begin{equation} \mathbf{[5]}\otimes\mathbf{[5]} = \mathbf{[10]_{A}}\oplus\mathbf{[15]_{S}} \end{equation} From there, one can immediately identify the anty-symmetric $\mathbf{[10]_{A}}$ as,

\begin{equation} \mathbf{[10]_{A}} = (\mathbf{\bar{3}},\mathbf{1})\oplus (\mathbf{3},\mathbf{2})_{A}\oplus(\mathbf{1},\mathbf{3})_{A} \end{equation}

From the representations of the SM field, The $SU(3)_{c}\times SU(2)_{L}$ quantum numbers $(a,b)$ for fermionic states are: \begin{align} \begin{pmatrix} \nu_e & e^- \\ \end{pmatrix}_{L} &: (\mathbf{1},\mathbf{2}) \equiv (\mathbf{1},\bar{\mathbf{2}}) \\ e^{+}_{L} &: (\mathbf{1},\mathbf{1}) \\ \begin{pmatrix} u^{i} & d^{i} \\ \end{pmatrix}_{L} &: (\mathbf{3},\mathbf{2}) \\ u_L^{ic} &: (\bar{\mathbf{3}},\mathbf{1}) \\ d_L^{ic} &: (\bar{\mathbf{3}},\mathbf{1}) \end{align} One can find how exactly they fit in $SU(5)$. \begin{equation} \mathrm{SM} = \mathbf{[\bar{5}]}\oplus\mathbf{[10]} \end{equation} Note. To see the (anti-)symmetric structure of the representations one has to play with Young tableaux.


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