Deducing Lorentz representation out of symmetry type

Question

Cross-posted from here

Lorentz algebra can be proven to be isomorphic to $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$, so every representation can be denoted by two indices or spins, $(j_1, j_2)$.

Let's separate a tensor with 2 indices $T_{\mu\nu}$ as the sum of tensor with definite symmetry:

$$ T_{\mu\nu} = \left\{T_{(\mu\nu)} - \frac{1}{D}g_{\mu\nu}T^\lambda_\lambda\right\} + T_{[\mu\nu]} + \frac{1}{D}g_{\mu\nu}T^\lambda_\lambda\ , \tag1$$

where the 1st term, $\{\cdots\}$, is symmetric and traceless (provided $g_{\mu\nu}g^{\mu\nu} = \delta^\lambda_\lambda = D$), the 2nd one is antisymmetric and the last one is the trace term. Clearly,

$$ T_{(\mu\nu)} = \frac{T_{\mu\nu} + T_{\nu\mu}}{2}\ , \mbox{ mutatis mutandis for the antisymmetric term }T_{[\mu\nu]} $$

Usually, it is written that decomposition in Eq. (1) can be interpeted as decomposition in spins such that

$$ (1/2, 1/2)\otimes(1/2, 1/2) = (1, 1) \oplus [(1, 0) \oplus (0, 1)] \oplus (0, 0) \tag2$$

I know how to calculate Eq. (2) starting from $T_{\mu\nu} \in (1/2, 1/2)\otimes(1/2, 1/2)$, but my question is: how can you deduce this spin decomposition by looking only to Eq. (1), i.e., by knowing the symmetry of each component of the tensor? Or in other words, if I give you some Lorentz tensor with some definite symmetry, how do you deduce what representation it belongs to?

Answer 1

A convenient way to specify a highest weight representations of a classical Lie algebra is by its associated Young tableau. Typically, these have rows which look like $\Box \hspace{-0.05cm} \Box \hspace{-0.05cm} \Box \hspace{-0.05cm} \Box$. Each row must be shorter than the row above it. The algebra you're talking about is the complexified $\mathfrak{so}(4)$. Since this has rank 2, its Young tableaux can have at most two rows. The number of boxes in a row denotes the number of indices that are symmetrized. So if you have 4 boxes on top and 2 on the bottom, you have two sets of indices which are separately symmetrized but anti-symmetrized with each other.

Let the number of boxes in row $i$ be $m_i$. There is a convenient basis, called the Cartan basis, which expresses the highest weight of the irrep by \begin{equation} w = m_1 e_1 + m_2 e_2. \end{equation} There is another convenient basis, called the Dynkin basis, which is defined by \begin{equation} m_1 = \frac{\lambda_1 + \lambda_2}{2}, \;\;\; m_2 = \frac{\lambda_1 - \lambda_2}{2} \end{equation} where the $\lambda_i$ are non-negative integers. Incidentally, this basis shows how the Young tableau formalism can be straightforwardly extended to spinor irreps. We just allow the $m_i$ to be half-integers. Another extension suggested by this is that we should not really say that the number of boxes in row 2 is $m_2$ but $|m_2|$ since $m_2$ can be negative.

Now the question is how this relates to $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$ which, as you noted, has highest weights given by a pair of spins \begin{equation} w = j_1 f_1 + j_2 f_2. \end{equation} Since the rank is 2, we need two non-trivial representations (called fundamental representations) to see how the isomorphism works. To make our lives easy, let the first be the spinor representation which is $(m_1, m_2) = \left ( \frac{1}{2}, \frac{1}{2} \right )$ in one basis and $(j_1, j_2) = (1, 0)$ in the other. Similarly, let the second be the conjugate spinor which is $(m_1, m_2) = \left ( \frac{1}{2}, -\frac{1}{2} \right )$ in one basis and $(j_1, j_2) = (0, 1)$ in the other. This lets us figure out the change of basis between $e_i$ and $f_i$ which is enough information to always be able to go between $\mathfrak{so}(4)$ Young tableaux and $\mathfrak{su}(2)$ spins. Incidentally, the answer turns out to be that the spins are half the Dynkin labels.