Inverse and Transpose of Lorentz Transformation

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:

$$\Lambda^{-1} = \eta \Lambda^T \eta$$

$\def\th{\theta} \def\sp{\hspace{1ex}} \def\b{\beta} \def\a{\alpha} \def\m{\mu} \def\n{\nu} \def\g{\gamma} \def\d{\delta} \def\mt{\eta} \def\mti{\mt^{-1}} \def\F{\Phi} \def\Ft{\widetilde{\F}} \def\L{\Lambda} \def\Li{\L^{-1}} \def\Lt{\L^T} \def\id{\mathbb{I}}$From the invariant interval on derives \begin{align*} \L^\b_{\sp\a} \mt_{\b\g} \L^\g_{\sp\d} &= \mt_{\a\d}.\tag{1} \end{align*} Let $\F_\a^{\sp\b}=\L^\b_{\sp\a}$, so (1) is written as \begin{align*} \F_\a^{\sp\b} \mt_{\b\g} \L^\g_{\sp\d} &= \mt_{\a\d}\tag{2} \end{align*} with matrix interpretation \begin{align*} \F\mt\L=\mt.\tag{3} \end{align*} By index gymnastics (2) is massaged into the form $$\F^\a_{\sp\g} \L^\g_{\sp\d} = \d^\a_\d$$ so $\F^\a_{\sp\b} = (\Li)^\a_{\sp\b}$. Note that, critically, the first index has been raised and the last lowered. Letting $\Ft$ be the matrix determined by $\F^\a_{\sp\b}$ we have $$\Ft=\Li.$$ We are, however, interested in $\F_\a^{\sp\b}$. We find $\F_\a^{\sp\b} = \mt_{\g\a}\F^\g_{\sp\d}\mt^{\b\d} = \mt_{\g\a}(\Li)^\g_{\sp\d}\mt^{\b\d}$. This has the matrix interpretation \begin{align*} \F &= \mt\Li\mti. \tag{4} \end{align*} In fact, (4) follows immediately from (3), illustrating the usefulness of the matrix representation. The confusion boils down to one between $\F$ and $\Ft$. It is straightforward to show from (4) and the general form for $\L$ that $\F=\Lt$. (See comment below.) Thus, $$\Lt\mt\L=\mt$$ is the correct matrix representation of (1).

Let us illustrate the difference between $\F$ and $\Ft$ with a specific nontrivial example. Represent $\L^\a_{\sp\b}$ by $$\L = \left[ \begin{array}{cccc} \gamma & 0 & 0 & -\beta \gamma \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ -\beta \gamma & 0 & 0 & \gamma \\ \end{array} \right].$$ Then $$\F = \Lt = \left[ \begin{array}{cccc} \gamma & 0 & 0 & -\beta \gamma \\ 0 & \cos \theta & \sin \theta & 0 \\ 0 & -\sin \theta & \cos \theta & 0 \\ -\beta \gamma & 0 & 0 & \gamma \\ \end{array} \right].$$ But $$\Ft = \Li = \left[ \begin{array}{cccc} \gamma & 0 & 0 & \beta \gamma \\ 0 & \cos \theta & \sin \theta & 0 \\ 0 & -\sin \theta & \cos \theta & 0 \\ \beta \gamma & 0 & 0 & \gamma \\ \end{array} \right].$$ (This is a boost in the $z$ direction and rotation about the $z$-axis.)

Comment

When one studies the general solutions to (1) one finds that they are combinations of rotations and boosts. Note that for a rotation $\mt \L(\th)^{-1}\mti=\mt\L(-\th)\mti=\L(-\th)=\L(\th)^T$ and that for a boost, $\mt \L(\b)^{-1}\mt=\mt \L(-\b)\mti= \L(\b)=\L(\b)^T$.

1. Note that the conventional definition of transposed matrix $$(\Lambda^T)_{\nu}{}^{\mu} ~:=~ \Lambda^{\mu}{}_{\nu}.\tag{1'} $$ is slightly different than OP's definition (1).

2. In plain English: When we don't apply the metric, the Lorentz matrix $\Lambda$ has conventionally NW-SE slanted indices, while the transposed matrix $\Lambda^T$ has SW-NE slanted indices.

   See also e.g. this & this related Phys.SE post.

3. Incidentally, OP's eq. (1) is consistent with eq. (1') after appropriately raising and lowing indices with the metric.

4. OP's eq. (D) is wrong because it doesn't comply with above convention.

5. More details: In matrix form OP's eqs. (A)-(C) read $$ \Lambda^{-1}~=~ (\eta\Lambda\eta^{-1})^T,\tag{A} $$ $$ \eta^{-1}\Lambda^T\eta~=~ (\eta\Lambda\eta^{-1})^T,\tag{B} $$ $$ \eta^{-1}\Lambda^T\eta~=~\Lambda^{-1},\tag{C} $$ respectively, which are indeed all true. Eq. (C) follows from the definition $$ \Lambda^T\eta\Lambda~=~\eta$$ of a Lorentz matrix.