# Tensor product of fundamental representation of $SU(5)$

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.

• Almost all standard textbooks on Lie algebra in particle physics discussed this thing. One can check Georgi or Zee's books.
– Mass
Oct 8, 2021 at 1:07
• 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. Oct 8, 2021 at 1:12
• Oct 8, 2021 at 12:25

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

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

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 $$\mathbf{}$$, we can write $$\begin{equation*} \mathbf{} \rightarrow \begin{pmatrix} SU(3)_{c} & \\ & SU(2)_{L} \\ \end{pmatrix} \begin{pmatrix} q_{i} \\ \ell_{r} \end{pmatrix} \end{equation*}$$ Recall that quarks are triplet under $$SU(3)_{c}$$ and leptons are color blind, which simply put in the following way, $$\begin{eqnarray} \mathbf{} & = & \mathbf{(3,1)}\oplus \mathbf{(1,2)} \\ \mathbf{[\bar{5}]} & = & \mathbf{(\bar{3},1)} \oplus \mathbf{(1,\bar{2})} \end{eqnarray}$$ 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. $$\begin{equation*} \bar{\mathbf{}} = (\mathbf{3},\mathbf{1}) \oplus (\mathbf{1},\mathbf{2}) = \begin{pmatrix} d^{rc} & d^{bc} & d^{gc} & e^{-} & -\nu_e \end{pmatrix}_L \end{equation*}$$ 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{}\otimes\mathbf{} &= [(\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{} \otimes \mathbf{} &= \mathbf{}_{S}\oplus\mathbf{[\bar{3}}_{A}]\\ \mathbf{} \otimes \mathbf{} &= \mathbf{}_{S}\oplus\mathbf{}_{A} \end{align}

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

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

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

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 $$SU(5)$$. $$\begin{equation} \mathrm{SM} = \mathbf{[\bar{5}]}\oplus\mathbf{} \end{equation}$$ Note. To see the (anti-)symmetric structure of the representations one has to play with Young tableaux.