# Decomposition of the torsion tensor

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.

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 $ef$.

• You need to reach to SL (2,C) spinors using the Infeld- van der Waerden symbols Jan 22, 2018 at 18:08
• @DanielC: Thanks for that tip. I will have a look at it, although I am afraid it is not going to be easy for me: I mean, the torsion tensor will in that notation have three dotted and three undotted indices, and thus any projection operators will in that notation carry twelve indices, I guess. Jan 23, 2018 at 7:38
• @DanielC: Shortly afterwards, it occured to me that if projection operators are possible using the notation you suggest, then it should also be possible without it, i.e., in the tensor notation (although now over the complex domain), because the conversion back and forth between the two formalisms is one-to-one, using essentially the identity matrix and the Pauli matrices. And I now believe to have identified the sought after projection operators, see my update above. Jan 26, 2018 at 7:04

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.