Decomposition of the torsion tensor

Question

See Update below.

Consider the torsion tensor $T_{\mu\nu\rho} = -T_{\mu\rho\nu}$. In a local Lorentz frame, as determined by a vierbein $e^{a}{}_{\mu}$, it may equivalently be given as $T_{abc} = e^{\mu}{}_{a}e^{\nu}{}_{b}e^{\rho}{}_{c}T_{\mu\nu\rho}$. Under Lorentz transformations, this third rank Lorentz tensor transforms in the \begin{align} (\frac{1}{2},\frac{1}{2}) \otimes ((1,0) \oplus(0,1)) = (\frac{1}{2},\frac{1}{2}) \oplus (\frac{3}{2},\frac{1}{2}) \oplus (\frac{1}{2},\frac{1}{2}) \oplus (\frac{1}{2},\frac{3}{2}) \end{align} Lorentz representation. The two $(\frac{1}{2},\frac{1}{2})$'s correspond to the vector and axial vector parts, respectively, given by $V^{def}_{abc}T_{def}$ and $A^{def}_{abc}T_{def}$, introducing the following projection operators (each manifestly antisymmetric in both $bc$ and $ef$): \begin{align} V^{def}_{abc} & \equiv \frac{1}{6}\eta^{de}(\eta_{ab}\delta^{f}_{c} - \eta_{ac}\delta^{f}_{b}) - \frac{1}{6}\eta^{df}(\eta_{ab}\delta^{e}_{c} - \eta_{ac}\delta^{e}_{b}), \\ A^{def}_{abc} & \equiv \frac{1}{6}\delta^{def}_{abc}, \end{align} where $\eta_{ab}$ is the Minkowski metric, and $\delta^{def}_{abc}$ is a generalized Kronecker delta. The $(\frac{3}{2},\frac{1}{2}) \oplus (\frac{1}{2},\frac{3}{2})$ part is given by $Q^{def}_{abc}T_{def} \equiv T_{abc} - (V^{def}_{abc} + A^{def}_{abc})T_{def}$, defining the projection operator $Q^{def}_{abc}$. I have checked that $V^{def}_{abc}$, $A^{def}_{abc}$, and $Q^{def}_{abc}$ in conjunction do form a proper projection algebra as concerns idempotency, orthogonality, and completeness. But my question is the following:

Question: How, if at all possible, can $(\frac{3}{2},\frac{1}{2})$ and $(\frac{1}{2},\frac{3}{2})$ be individually projected out? What, if any, are the projection operators? If possible, does it require going to the complex domain, in analogy with the decomposition of, say, the Faraday tensor into the self-dual and anti-self-dual parts $(1,0)$ and $(0,1)$?

I am asking because I am unable to identity the form of any such projection operators. I have tried using the Levi-Civita tensor to construct some self-dual and anti-self-dual projection operators (over the complex domain), doubly contracting on the last two indices of $Q^{def}_{abc}T_{def}$, but the resulting operators are not even individually idempotent.

PS: Any relevant or helpful link will of course be appreciated.

Update: I think I have solved it myself. I belive the sought after projection operators, $(Q_{\pm})^{def}_{abc}$ say, projecting out $(\frac{3}{2},\frac{1}{2})$ and $(\frac{1}{2},\frac{3}{2})$, respectively, are given by

\begin{align} (Q_{\pm})^{def}_{abc} = \frac{1}{2}Q^{def}_{abc} \pm (D_{1})^{def}_{abc} \pm (D_{2})^{def}_{abc}, \end{align}

correlated signs, where

\begin{align} (D_{1})^{def}_{abc} &\equiv \frac{i}{6}\varepsilon_{bc}{}^{ef}\delta^{d}_{a}, \\ (D_{2})^{def}_{abc} &\equiv \frac{i}{12}(\varepsilon_{ab}{}^{de}\delta^{f}_{c} - \varepsilon_{ac}{}^{de}\delta^{f}_{b} - \varepsilon_{ab}{}^{df}\delta^{e}_{c} + \varepsilon_{ac}{}^{df}\delta^{e}_{b}). \end{align}

As for the other projection operators in this post, these projection operators have been written in a form in which they are manifestly antisymmetric in both $bc$ and $ef$.

Answer 1

Maybe I don't understand clearly your query, but I think it's pretty easy to decompose the torsion tensor. Since $T_{cab} = -\: T_{cba}$, you get $4 \times 6 = 24$ independant components in 4D space-time. You can extract (or define) two independant vectors which gives you 8 components (the global constant is partly arbitrary. I'm using the $\eta = (1, -1, -1, -1)$ convention) : \begin{align} v_a &\equiv -\: \frac{1}{3} \: \eta^{bc} \, T_{cab}, \quad \text{vectorial part} \tag{1} \\[12pt] \tau^a &\equiv \frac{1}{3!} \: \varepsilon^{abcd} \, T_{bcd}. \quad \text{pseudo-vectorial or axial part} \tag{2} \end{align} Then there are 16 components remaining, grouped into a traceless tensor $\mathcal{T}$ : \begin{align} T_{cab} &= T_{cab}^{\mathrm{V}} + T_{cab}^{\mathrm{A}} + T_{cab}^{\mathrm{T}} \\[12pt] &= (\eta_{ac} \, v_b - \eta_{bc} \, v_a) + \varepsilon_{abcd} \, \tau^d + \mathcal{T}_{cab}. \tag{3} \end{align} Most authors are assuming that the traceless tensorial part is vanishing : $\mathcal{T}_{cab} = 0$, but it isn't in the most general case.