I've seen this question asked a few times on Stack Exchange, but I'm still quite confused why the following "contradiction" seems to arise.
By definition:
- $(\Lambda^T)^{\mu}{}_{\nu} = \Lambda_{\nu}{}^{\mu}$
- $\Lambda^T \eta \Lambda = \eta$, which is $\Lambda^{\rho}{}_{\mu} \eta_{\rho \sigma} \Lambda^{\sigma}{}_{\nu} = \eta_{\mu \nu}$ in index notation.
We can further manipulate the second definition (as done in Tong's lecture notes):
$\begin{align}
\Lambda^{\rho}{}_{\mu} \eta_{\rho \sigma} \Lambda^{\sigma}{}_{\nu} &= \eta_{\mu \nu} \\
\Lambda^{\rho}{}_{\mu} \Lambda_{\rho \nu} &= \eta_{\mu \nu} \\
\Lambda^{\rho}{}_{\mu} \Lambda_{\rho \nu} \eta^{\nu \sigma} &= \eta_{\mu \nu} \eta^{\nu \sigma} \\ \Lambda^{\rho}{}_{\mu} \Lambda_{\rho}{}^{\sigma} &= \delta_{\mu}^{\sigma} \\ \Lambda_{\rho}{}^{\sigma} \Lambda^{\rho}{}_{\mu} &= \delta_{\mu}^{\sigma}
\end{align}$
Recalling that $({\Lambda^{-1}})^{\sigma}{}_{\rho}$ is defined through:
$$({\Lambda^{-1}})^{\sigma}{}_{\rho} \Lambda^{\rho}{}_{\mu} = \delta^{\sigma}_{\mu}$$
This then implies that $$({\Lambda^{-1}})^{\sigma}{}_{\rho} = \Lambda_{\rho}{}^{\sigma}\tag{A}$$ but according to definition 1, doesn't $$({\Lambda^{T}})^{\sigma}{}_{\rho} = \Lambda_{\rho}{}^{\sigma}~?\tag{B}$$ This seems to incorrectly imply that $$({\Lambda^{T}})^{\sigma}{}_{\rho} = ({\Lambda^{-1}})^{\sigma}{}_{\rho}\tag{C}.$$ I'm not really sure what step of my logic is incorrect.
Tong makes the following comment on the (A) result:
The result is analogous to the statement that the inverse of a rotation matrix is the transpose matrix. For general Lorentz transformations, we learn that the inverse is sort of the transpose where “sort of” means that there are minus signs from raising and lowering. The placement of indices in tells us where those minus signs go.
This comment seems to suggest that (B) is incorrect – although it just seems like mere application of definition 1.
Edit to clarify question after initial responses:
From this analysis, why is it incorrect to conclude that $$\Lambda^{-1} = \Lambda^T~?\tag{D}$$ We know this matrix equation is not true, but why is this not implied by $({\Lambda^{T}})^{\sigma}{}_{\rho} = \Lambda_{\rho}{}^{\sigma} = ({\Lambda^{-1}})^{\sigma}{}_{\rho}$ since the indices in $\Lambda^{T}$ and $\Lambda^{-1}$ are the same?
Further Clarification Question:
Some of the responses will reveal that in fact only the matrix equation D is incorrect because the index structure of $\Lambda$ is $\Lambda^{\mu}{}_{\nu}$, the index structure of $\Lambda^{-1}$ is ${(\Lambda^{-1})}^{\mu}{}_{\nu}$, but the index structure of $\Lambda^T$ is ${(\Lambda^{T})}_{\mu}{}^{\nu}$ (not ${(\Lambda^{T})}^{\mu}{}_{\nu}$).
However, this leaves one final question: how can we show explicitly that the matrix $\Lambda^T$ should correspond to this different index structure? Using this structure does make everything consistent again, but how does this follow from defining the matrix $\Lambda$ as corresponding to $\Lambda^{\mu}{}_{\nu}$?
Best Answer
After a very helpful discussion in the comments section and reading the responses, I thought I would type up (from my perspective) what I have learnt in case it will help anyone with the same question.
$$({\Lambda^{T}})^{\sigma}{}_{\rho} = \Lambda_{\rho}{}^{\sigma} = ({\Lambda^{-1}})^{\sigma}{}_{\rho}$$
is in fact a correct statement, but we have to be careful when converting this back into a matrix equation.
We should interpret $\Lambda$ as $\Lambda^{\mu}{}_{\nu}$, $\Lambda^{-1}$ as $({\Lambda^{-1}})^{\mu}{}_{\nu}$, but $\Lambda^T$ should be interpreted as $(\Lambda^T)_{\mu}{}^{\nu}$.
Therefore, we cannot interpret $({\Lambda^{T}})^{\sigma}{}_{\rho}$ as $\Lambda^T$ so equation D is incorrect. Instead, using the metric, $({\Lambda^{T}})^{\sigma}{}_{\rho} = \eta^{\sigma \alpha} (\Lambda^T)_{\alpha}{}^{\beta} \eta_{\beta \rho}$. So, instead of matrix equation D, we really should have:
$$\Lambda^{-1} = \eta \Lambda^T \eta$$