In Srednicki's textbook Quantum Field Theory, section 97 discusses Grand Unification. On page 606, it states:

In terms of $\rm SU(5)$, we have \begin{equation} 5 \otimes 5 = 15_{S} \oplus 10_{A} \tag{97.5} \end{equation}

where the subscripts $S$ and $A$ refer to symmetric and antisymmetric respectively. To my understanding, $15_{S}$ is a $15 \times 15$ matrix, and $10_{A}$ is a $10 \times 10$ matrix. Am I right? However, in the text, a left-handed Weyl field $\chi_{ij} = - \chi_{ji}$ in the 10 representation is defined. Its components are given by \begin{equation} \chi_{ij} = \left( \begin{array}{ccccc} 0 & \overline{u}^{g} & -\overline{u}^{b} & u_{r} & d_{r} \\ -\overline{u}^{g} & 0 & \overline{u}^{r} & u_{b} & d_{b} \\ \overline{u}^{b} & -\overline{u}^{r} & 0 & u_{g} & d_{g} \\ -u_{r} & - u_{b} & -u_{g} & 0 & \overline{e} \\ -d_{r} & -d_{b} & -d_{g} & -\overline{e} & 0 \end{array} \right). \tag{97.12} \end{equation} Why is $\chi_{ij}$ not a $10 \times 10$ matrix, but a $5\times 5$ matrix?


Representations are vector spaces that behave a certain way when the group acts on them. Representations are often labelled by their vector space dimension. The fundamental representation of $SU(5)$ is just the space of 5-dimensional complex vectors, so we call it $5$. The representation $10_A$ is a 10-dimensional vector space, and $15_S$ is a 15-dimensional vector space.

Now, it so happens that one convenient way of representing the vectors of $10_A$ is as anti-symmetric $5\times 5$ matrices. That is, the 10 basis "vectors" of $10_A$ are the matrices $\chi^{12}, \chi^{13}, \dots$. where $$\chi^{12} = \begin{pmatrix}0 & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 &0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{pmatrix}$$ and so on.

You can work out that there are "5 choose 2" = 10 of these basis vectors. Likewise, you can represent the 15 basis vectors of $15_S$ as the 15 independent symmetric $5\times 5$ matrices.

The key thing to remember is that the vectors in a representation are an abstract thing. You can represent them as lists of numbers, or as particular matrices, or as anything else.


Am I right?

Only in a small way, but basically not. (97.5) denotes dimensionalities of the irreducible representations of SU(5) involved, so how the respective vectors are acted upon by the coproduct of SU(5) generators.

On the left hand side, you have two quintuplets (5-vectors), each acted upon by 5×5 matrices $T^a_5$, so the whole reducible rep acted upon by 25×25 matrices $$ \Delta (T^a)_{25}= T^a_5\otimes 1\!\!1 _5 + 1\!\!1 _5 \otimes T^a_5 . $$ On the right hand side, you have the reduction of the 25-dim vector into two separated vectors of dimension 15 and 10 respectively, each one acted on by 15×15 and 10×10 generators respectively, $$ T^a_{15} \oplus T^a_{10} , $$ That is the 25×25 matrices break up consistently into upper left 15×15 blocks and lower right 10×10 blocks, as your group theory text should detail. All 5,25,15,10 dim matrices obey the very same su(5) Lie algebra!

So (97.12) is basically a 10-dim complex vector, with the 10 degrees of freedom of the upper triangular piece arranged into a 5×5 antisymmetric matrix format by superfluous replication for future convenience. (It couples to a fermion 5 and a Higgs 5 in the invariant term in the lagrangian, so it pays to have two loose 5 indices to saturate, instead of a loose 10.)

An analog you might use to fix your thinking is looking at the Kronecker composition of two doublets (spin 1/2s) of SU(2) into a (symmetric) triplet (spin one) and an antisymmetric singlet (spin 0), $$ 2\otimes 2= 3_s\oplus 1_A . $$ If you chose to, you could arrange the 3-vector on the right into the format of a real antisymmetric 3×3 matrix, as one routinely does in the logarithm of 3×3 rotation matrices, which amounts to three angular momentum generators $\vec L$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.