Determinant of rank-2 tensor expression

Question

I have some tensorial expression from Broderick, A. and Blandford, R. 2003 (e.q. 76), which reads,

$$ \Omega^{\mu}_{\nu} = \alpha \delta^{\mu}_{\nu} - i \gamma M^{\mu}_{\nu}$$

where $M_{\mu \nu} = -M_{\nu \mu} = \epsilon_{\mu \nu \alpha \beta} u^{\alpha} B^{\beta}$ and the indexes run $\mu,\nu = 0,1,2,3$

How do I find the determinant of this expression?

In the paper, it is given that

$$ \text{det } \Omega^{\mu}_{\nu} = \alpha^4 - \alpha^2 \gamma^2 B^{\mu}B_{\mu}$$

but I am clueless as to how this expression was arrived at.

Any help would be greatly appreciated. Thank you

Answer 1

I shall lower the index and consider $\Omega_{ab}$ where $a,b=0,1,2,3$. The determinant is given by $\det(\Omega_{\mu\nu})=\epsilon^{abcd}\,\Omega_{0a}\,\Omega_{1b}\,\Omega_{2c}\,\Omega_{3d}$ after lowering the indices. You can then just plug in the $\Omega_{ab}$ and use the properties of $\delta_{ab}$ and $\epsilon^{abcd}$. This is rather a brute force approach (and tedious!), so I don't know if there is a quicker, more elegant way to do it. I also notice a $u^{a}$ in the definition of $M_{ab}$; I presume there is some normalization $(u^{a}u_{a}= -1$ perhaps?) that will need to be taken into account also.