# Determinant of the inverse of a Lorentz transformation

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.

• Jul 18, 2022 at 6:54
• $\Lambda^T \ne \Lambda^{-1}$ Jul 18, 2022 at 7:34
• I think it is best to not mix matrix notation and Einstein notation. The matrix notation is unnecessary and less expressive in general
– Dale
Jul 18, 2022 at 11:53

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 $$\det ( \Lambda^T ) = \det(\eta) \det(\Lambda^{-1}) \det(\eta^{-1}) \quad \implies \quad \det \Lambda = \det ( \Lambda^{-1}) .$$
• Doesn't$$\forall\mu:\forall\nu:(\Lambda^T)_\mu{}^\nu = (\Lambda^{-1})_\mu{}^\nu$$mean that $\Lambda^T = \Lambda^{-1}$? Jul 19, 2022 at 10:25
• No, because $\Lambda^{-1}$ and$\Lambda^T$ have different intrinsic index structures. Jul 19, 2022 at 10:27
• I mentioned it in the first line but let me elaborate. The matrices $\Lambda$ and $\Lambda^{-1}$ naturally have its first index up and second index down whereas $\Lambda^T$ has its first index down and second one up. So when I write $\Lambda_\mu{}^\nu$ that does NOT represent a Lorentz matrix because its first index is NOT up and second one is NOT down. We must instead use the metric $\eta_{\mu\nu}$ to raise/lower indices to bring it into proper form (1/2) Jul 19, 2022 at 10:48
• so we can write $\Lambda_\mu{}^\nu = \eta_{\mu\rho} \eta^{\nu\sigma} \Lambda^\rho{}_\sigma$ and now the indices on $\Lambda$ has the right form. The same thing applies to $\Lambda^{-1}$ as I've done in the answer (2/2). Jul 19, 2022 at 10:49
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 $$\det A = \det A^{t}$$.