Dirac field in topologically massive gravity

Abstract

We consider a Dirac field coupled minimally to the Mielke–Baekler model of gravity and investigate cosmological solutions in three dimensions. We arrive at a family of solutions which exists even in the limit of vanishing cosmological constant.

Appendix: Irreducible decompositions

In this section we give briefly the irreducible pieces of torsion, contortion and curvature in three dimensions. Firstly torsion which has nine components can be decomposed

$$\begin{aligned} \underbrace{T^a}_{\# 9} = \underbrace{ {}^{(1)}T^a}_{\# 5} + \underbrace{{}^{(2)}T^a}_{\# 3} + \underbrace{{}^{(3)}T^a}_{\# 1} \end{aligned}$$

(39)

where ${}^{(2)}T^a = - \frac{1}{2} (\iota _b T^b) \wedge e^a, {}^{(3)}T^a = \frac{1}{3} \iota ^a (e_b \wedge T^b)$ and ${}^{(1)}T^a = T^a - {}^{(2)}T^a - {}^{(3)}T^a$. In this section the notation with the number under a brace is for the number of components of that part. They have the properties, ${}^{(1)}T^a \wedge e_a = {}^{(2)}T^a \wedge e_a =0$ and $\iota _a {}^{(1)}T^a = \iota _a {}^{(3)}T^a = 0$. Our choice (25) corresponds to

$$\begin{aligned} ^{(1)}T^{a}=0, \ \ ^{(2)}T^{a}= u\left(\begin{array}{c} e^{01}\\ 0\\ -e^{12} \\ \end{array} \right), \ \ ^{(3)}T^{a}=v\left(\begin{array}{c} e^{12}\\ e^{02}\\ -e^{01} \\ \end{array} \right) \end{aligned}$$

(40)

which means $4=3\oplus 1$. After the solution (32) we are left only with ${}^{(3)}T^{a}$.

Secondly one can decompose the contortion having nine components

$$\begin{aligned} \underbrace{K_{ab}}_{\# 9}=\underbrace{{}^{(1)}K_{ab}}_{\# 5} + \underbrace{{}^{(2)}K_{ab}}_{\# 3} + \underbrace{{}^{(3)}K_{ab}}_{\#1} \end{aligned}$$

(41)

where ${}^{(2)}K_{ab}=\frac{1}{2}[e_a \wedge (\iota ^cK_{cb}) - e_b \wedge (\iota ^cK_{ca})], {}^{(3)}K_{ab}=-\frac{1}{6}\iota _{ab} (K_{cd}\wedge e^{cd})$ and ${}^{(1)}K_{ab} = K_{ab} - {}^{(2)}K_{ab} - {}^{(3)}K_{ab}$. They have the properties $\iota _a{}^{(1)}K^{ab}=\iota _a{}^{(3)}K^{ab}=0$ and ${}^{(1)}K_{ab} \wedge e^{ab} = {}^{(2)}K_{ab} \wedge e^{ab}=0$. For our case (26) we possess again ${}^{(1)}K_{ab}=0$, but nonzero ${}^{(2)}K_{ab}$ and ${}^{(3)}K_{ab}$, i.e. $4=3\oplus 1$. Besides after the solution (32) only ${}^{(3)}K_{ab}$ survives.

Finally one can split the curvature with nine components

$$\begin{aligned} \underbrace{R_{ab}}_{\# 9}=\underbrace{{}^{(1)}R_{ab}}_{\# 5} + \underbrace{{}^{(2)}R_{ab}}_{\# 3} + \underbrace{{}^{(3)}R_{ab}}_{\#1} \end{aligned}$$

(42)

where ${}^{(2)}R_{ab}=\frac{1}{2}(e_a \wedge \iota _b - e_b \wedge \iota _a)(e^c \wedge R_c), {}^{(3)}R_{ab}=\frac{1}{6} R e_{ab}$ and ${}^{(1)}R_{ab} = R_{ab} - {}^{(2)}R_{ab} - {}^{(3)}R_{ab}$ with $R_a = \iota ^bR_{ba}$ and $R=\iota ^aR_a$. They have the properties $\iota _{ab}{}^{(1)}R^{ab}=\iota _{ab}{}^{(2)}R^{ab}=0, {}^{(1)}R_{ab} \wedge e^b = {}^{(3)}R_{ab} \wedge e^b=0$ and $e_b \wedge \iota _a {}^{(1)}R^{ab}=0$. For the solution (32) although all three pieces of curvature and of torsion are nonzero, since $DT^a =0$ the Ricci tensor is symmetric.