Determinant of the inverse of a Lorentz transformation

Question

In many text book (Ashok Das Quantum Field Theory) $$(\Lambda^T)_\nu{}^\mu=\Lambda^\mu{}_\nu$$ that gives $\Lambda^T$ = $\Lambda^{-1}$, where $\Lambda$ is Lorentz Transformation matrix. However, this is not general definition of transpose where rows and columns are exchanged. ie $\Lambda^i_j = \Lambda^j_i$. Hence element of transpose is not obtained by exchanging the rows and column. It is also given that $det(\Lambda^T) = det(\Lambda)$. However generally this is true when we take transpose (i.e rows and columns atre exchanged which not the case here). Is this just accidental that $det(\Lambda^{-1}) = det(\Lambda)$? or tru for any mixed rank tensor.

Answer 1

The correct index structure of the matrices are $\Lambda^\mu{}_\nu$, $(\Lambda^{-1})^\mu{}_\nu$ and $(\Lambda^T)_\mu{}^\nu$ and the correct formulas are $$ \Lambda^\nu{}_\mu = (\Lambda^T)_\mu{}^\nu = (\Lambda^{-1})_\mu{}^\nu. $$ The last equality does NOT imply that in matrix notation $\Lambda^T = \Lambda^{-1}$ because their index structures are different. Instead, we should write $$ (\Lambda^T)_\mu{}^\nu = (\Lambda^{-1})_\mu{}^\nu = \eta_{\mu\rho} \eta^{\nu\sigma}(\Lambda^{-1})^\rho{}_\sigma . $$ We can now employ matrix notation and write $$ \Lambda^T = \eta \Lambda^{-1} \eta^{-1} = \eta \Lambda^{-1} \eta . $$ We can now take the determinant on both sides and we find $$ \det ( \Lambda^T ) = \det(\eta) \det(\Lambda^{-1}) \det(\eta^{-1}) \quad \implies \quad \det \Lambda = \det ( \Lambda^{-1}) . $$

Answer 2

If $\Lambda$ is a Lorentz matrix, then its inverse is $$\Lambda^{-1}= \eta \Lambda^t \eta$$ where the transposed matrix has here the standard meaning and $\eta= diag(-1,1,1,1)$. Taking the determinant on both sides, it holds, $\det \Lambda^{-1} = \det\Lambda^{t} (= \det \Lambda)$ because $\det\eta=-1$. This fact is false in general but it is true for the elements of some matrix group like $O(n)$ and $Sp(n,R)$ in addition to the Lorentz group.

I finally stress that some groups of physical interest are such that the determinant of their matrices is $1$ by definition (think of the group of proper rotation or the special Lorentz group). In that case, the identity we are discussing is obvious since $\det A = \det A^{t}$.