Lorentz representation of totally symmetric tensors?

Question

Pretty much the title. A rank 1 tensor is $(1/2,1/2)$, a symmetric rank 2 tensor is $(0,0)+(1,1)$. I'm curious how this generalizes to rank $n$ totally symmetric tensors.

Answer 1

TL;DR: There is already a correct answer by MannyC. The totally symmetric rank-$n$ tensor product of a 4-vector representation decomposes as $$\begin{align}\begin{pmatrix}{\bf n+3} \cr{\bf 3}\end{pmatrix} ~=~&{\bf 4}^{\odot n}~=~(\underbrace{{\bf 2}_L}_{j_L=1/2}\otimes \underbrace{{\bf 2}_R}_{j_R=1/2})^{\odot n}\cr ~=~&\bigoplus_{k=0}^{\lfloor\frac{n}{2}\rfloor} \underbrace{({\bf n+1-2k})_L}_{j_L=n/2-k} \otimes \underbrace{({\bf n+1-2k})_R}_{j_R=n/2-k},\end{align}\tag{1}$$ cf. e.g. this & this Phys.SE posts.

Longer explanation:

1. Let $$V_L~\cong~\mathbb{C}^2\qquad\text{and}\qquad V_R~\cong~\mathbb{C}^2\tag{2}$$ denote the left and right Weyl spinor representations, respectively, of the double cover $G=SL(2,\mathbb{C})$ of the restricted Lorentz group $SO^+(1,3;\mathbb{R})$. Let $(e_{\alpha})_{\alpha=1,2}$ and $(e_{\dot{\alpha}})_{\dot{\alpha}=\dot{1},\dot{2}}$ denote bases for the vector spaces (2), respectively. The vector spaces (2) are endowed with antisymmetric forms, i.e. the Levi-Civita symbols $\epsilon_{\alpha\beta}$ and $\epsilon_{\dot{\alpha}\dot{\beta}}$, respectively.

2. The tensor product $$V=V_L\otimes_{\mathbb{C}} V_R~\cong~\mathbb{C}^{4} \tag{3}$$ is isometric to the complexified Minkowski$^1$ space, with metric $$\eta_{\mu\nu}~\sim~ \eta_{\alpha\dot{\alpha},\beta\dot{\beta}} ~=~\epsilon_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}.\tag{4}$$ The vector space $V$ has a basis $$(e_{\alpha}\otimes e_{\dot{\alpha}})_{\alpha=1,2;\dot{\alpha}=\dot{1},\dot{2}}.\tag{5}$$

3. The $(j_L,j_R)$ irreducible representation of $G$ is $$ V_L^{\odot 2j_L}\otimes_{\mathbb{C}}V_R^{\odot 2j_R}, \qquad j_L, j_R~\in~ \frac{1}{2}\mathbb{N}_0, \tag{6} $$ with basis $$ \left(e_1^{\odot m_L}\odot e_2^{\odot (2j_L- m_L)}\otimes e_{\dot{1}}^{\odot m_R}\odot e_{\dot{2}}^{\odot (2j_R- m_R)}\right)_{0\leq m_L\leq 2j_L;0\leq m_R\leq 2j_R}, \tag{7} $$ and dimension $(2j_L+1)(2j_R+1)$, cf. e.g. this Phys.SE post.

4. The symmetric tensor product $$V^{\odot n}, \qquad n~\in~ \mathbb{N}_0,\tag{8}$$ has dimension $ \binom{n+3}{3}, $ and corresponds to a symmetric tensor field $S^{\mu_1 \ldots\mu_{n}}.$

5. Similarly, a $n$th symmetric tensor field $$\eta_{\mu_1\mu_2}\ldots\eta_{\mu_{2k-1}\mu_{2k}} S^{\mu_1 \ldots\mu_{n}}\tag{9}$$ with $k\in \mathbb{N}_0$ traces corresponds to $V^{\odot (n-2k)}$ with dimension $ \binom{n-2k+3}{3}. $

6. We can split $$\begin{align} V^{\odot n}~\cong~&\underbrace{V^{\odot (n-2)}}_{\text{single-trace}}+\underbrace{ V_L^{\odot n}\otimes V_R^{\odot n}}_{\text{traceless}}, \cr \binom{n+3}{3}~=~&\binom{n+1}{3}+(n+1)^2 .\end{align}\tag{10} $$ Intuitively, eq. (10) can be understood as follows: The left/undotted and right/dotted bases commute in the irrep $V_L^{\odot n}\otimes V_R^{\odot n}$ but not in $V^{\odot n}$. The non-commutativity in $V^{\odot 2}$ is associated with pairs of non-zero trace.

7. Repeated applications of eq. (10) lead to the sought-for result (1) $$ V^{\odot n}~\cong~\bigoplus_{k=0}^{\lfloor \frac{n}{2}\rfloor} \underbrace{V_L^{\odot (n-2k)}}_{j_L=n/2-k}\otimes \underbrace{V_R^{\odot (n-2k)}}_{j_R=n/2-k}, \tag{11}$$ cf. e.g. this related Phys.SE post.

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$^1$ The word 'Minkowski' is here only a semantic label since the Minkowski signature is irrelevant in the complexification $\mathbb{C}^4$.

Answer 2

A symmetric traceless tensor of rank $\ell$ is $(\ell/2,\ell/2)$. If you want the reducible representation corresponding to a symmetric tensor, you have to add back the traces, which are lower rank tensors. So the representation is $$ \bigoplus_{k \equiv \,\ell \;\mathrm{mod}\;2}^\ell\, (k/2,k/2)\,. $$ Every trace reduces the rank by two, so the starting point of the sum is $0$ (1) if $\ell$ is even (odd).