Determinant of Rank-2 Tensor using Levi-Civita notation

Question

In my Professor's notes on Special Relativity, the determinant of a rank-two tensor $[T]$ (a $4\times 4$ matrix, basically) is given using the Levi Civita Symbol as: $$T=-\epsilon_{\mu\nu\rho\lambda}T^{\mu0}T^{\nu1}T^{\rho2}T^{\lambda3}.$$ But, while trying to derive this, I seem to be missing the minus sign. My attempt went as follows: I have a matrix of the form :

\begin{bmatrix} T^{00} & T^{01} & T^{02} & T^{03} \\ T^{10} & T^{11} & T^{12} & T^{13} \\ T^{20} & T^{21} & T^{22} & T^{23} \\ T^{30} & T^{31} & T^{32} & T^{33} \end{bmatrix}

As usual, by definition of the determinant of this matrix, I can use the four-dimensional Levi-Civita Symbol (with $\epsilon_{0123}=1$) to write it as: $$T=T^{00}\epsilon_{0\nu\rho\lambda}T^{\nu1}T^{\rho2}T^{\lambda3}-T^{10}\epsilon_{\nu1\rho\lambda}T^{\nu1}T^{\rho2}T^{\lambda3}+T^{20}\epsilon_{\nu\rho2\lambda}T^{\nu1}T^{\rho2}T^{\lambda3}-T^{30}\epsilon_{\nu\rho\lambda3}T^{\nu1}T^{\rho2}T^{\lambda3}.$$ In the first term the greek indices take spatial values{1,2,3}, while in the second term they take {0,2,3} and so on, and repeated indices imply summation. Now, using the even-odd permutation property of the Levi-Civita symbol, I have that $\epsilon_{\nu1\rho\lambda}=-\epsilon_{1\nu\rho\lambda}$ and $\epsilon_{\nu\rho\lambda3}=-\epsilon_{3\nu\rho\lambda}$. Plugging these into the former equation and using the summation convention by introducing the dummy index $\mu$, I find that:

$$T=\epsilon_{\mu\nu\rho\lambda}T^{\mu0}T^{\nu1}T^{\rho2}T^{\lambda3}. $$

Could someone please point out what went wrong in the derivation?

Answer 1

Levi-cevita symbol which is defined as $$ \varepsilon_{\mu\nu\rho\sigma}= \begin{cases} +1&\text{if}\,{\mu\nu\rho\sigma} \rm \, is\,an\,even\,permutation\,of\,0123 \\ -1&\text{if}\, {\mu\nu\rho\sigma}\rm \,is\,an\,odd\,permutation\,of\,0123\\ 0&\text{otherwise} \end{cases} $$ is a density tensor with weight $w=-1$; Based on definition of determinant we can write determinant of a matrix as $$\text{Det}(T) = \varepsilon_{\mu\nu\rho\sigma} T^{\mu}_{0} T^{\nu}_{1} T^{\rho}_{2} T^{\sigma}_{3} $$ or $$\text{Det}(T) = \varepsilon^{\mu\nu\rho\sigma} T_{\mu}^{ 0} T_{\nu}^{ 1} T_{\rho}^{ 2} T_{\sigma}^{ 3} $$ Also we can construct an absolute tensor for each density tensor, that for Levi-Civita density tensor is as $$\epsilon_{\mu\nu\rho\sigma}=\sqrt{\left|g\right|}\varepsilon_{\mu\nu\rho\sigma}$$
$$\epsilon^{\mu\nu\rho\sigma}=\frac{\text{sgn}(g)}{\sqrt{\left|g\right|}}\varepsilon_{\mu\nu\rho\sigma}$$
Then for Minkophski spacetime we have: $$\epsilon_{\mu\nu\rho\sigma}= \varepsilon_{\mu\nu\rho\sigma}$$ $$\epsilon^{\mu\nu\rho\sigma}= - \varepsilon^{\mu\nu\rho\sigma}$$ Now if we try to write determinant of $T$ as Contraction of two tensor we have

$$\text{Det}(T) = \varepsilon^{\mu\nu\rho\sigma} T_{\mu}^{ 0} T_{\nu}^{ 1} T_{\rho}^{ 2} T_{\sigma}^{ 3} = -\epsilon^{\mu\nu\rho\sigma} T_{\mu}^{ 0} T_{\nu}^{ 1} T_{\rho}^{ 2} T_{\sigma}^{ 3} = -\epsilon_{\mu\nu\rho\sigma} T^{\mu 0} T^{\nu 1} T^{\rho 2} T^{\sigma 3 }$$ Therefore your professor was right.

Answer 2

Not all rank 2 tensors are the same. In particular, this would make sense if $\mathbf T$ is meant to be a $(1,1)$-tensor with components $T^\mu_{\ \ \ \nu}$. If that's the case, then there would be an extra minus sign which comes from raising the second index prior to contracting with the Levi-Civita symbol.